4 APPLICATIONS OF DIFFERENTIATION

Transcription

1 4 APPLICATIONS OF DIFFERENTIATION 4. Maimum and Minimum Values. Afunctionf has an absolute minimum at c if f(c) is the smallest function value on the entire domain of f, whereas f has a local minimum at c if f(c) is the smallest function value when is near c.. Absolute maimum at s, absolute minimum at r,localmaimumatc, local minima at b and r, neither a maimum nor a minimum at a and d. 5. Absolute maimum value is f(4) 5; there is no absolute minimum value; local maimum values are f(4) 5 and f(6) 4; local minimum values are f() and f() f(5). 7. Absolute minimum at, absolute maimum at, local minimum at 4 9. Absolute maimum at 5, absolute minimum at, local maimum at, local minima at and 4. (a) (b) (c). (a) Note: By the Etreme Value Theorem, f must not be continuous; because if it were, it would attain an absolute minimum. (b)

2 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION 5. f() 8,. Absolute maimum f() 5; no local maimum. No absolute or local minimum. 7. f(), 0 <<. No absolute or local maimum or minimum value. 9. f(), 0 <. Absolute minimum f(0) 0; no local minimum. No absolute or local maimum.. f(),. Absolute maimum f( ) 9. No local maimum. Absolute and local minimum f(0) 0.. f(t) /t, 0 <t<. No maimum or minimum. 5. f(). Absolute maimum f(0) ;nolocal maimum. No absolute or local minimum. 7. f() if 0 < 4 if Absolute maimum f() ; no local maimum. No absolute or local minimum.

3 SECTION 4. MAXIMUM AND MINIMUM VALUES 9. f() 5 +4 f 0 () f 0 () 0,so 5 5 is the only critical number.. f() + 4 f 0 () +6 4 ( + 8). f 0 () 0 ( +4)( ) 0 4,. These are the only critical numbers.. s(t) t 4 +4t 6t s 0 (t) t +t t. s 0 (t) 0 t(t + t ) t 0 or t + t 0. Using the quadratic formula to solve the latter equation gives us t ± 4()( ) () ± ,.68. The three critical numbers are 0, ± g(y) y y y + g 0 (y) (y y +)() (y )(y ) y y + (y y +) y +y y( y) (y y +) (y y +) (y y +) (y y +). g 0 (y) 0 y 0,. The epression y y +is never equal to 0,sog 0 (y) eists for all real numbers. The critical numbers are 0 and. 7. h(t) t /4 t /4 h 0 (t) 4 t /4 4 t /4 4 t /4 (t / ) t 4 4 t. h 0 (t) 0 t t t 4 9. h0 (t) does not eist at t 0, so the critical numbers are 0 and F () 4/5 ( 4) F 0 () 4/5 ( 4) + ( 4) 4 5 /5 5 /5 ( 4)[5 +( 4) 4] ( 4)(4 6) ( 4)(7 8) 5 /5 5 /5 F 0 () 0 4, 8 7. F 0 (0) does not eist. Thus, the three critical numbers are 0, 8 7,and4. 4. f(θ) cosθ +sin θ f 0 (θ) sinθ +sinθ cos θ. f 0 (θ) 0 sinθ (cos θ ) 0 sin θ 0 or cos θ θ nπ [n an integer] or θ nπ. The solutions θ nπ include the solutions θ nπ, so the critical numbers are θ nπ. 4. A graph of f 0 0 sin () + is shown. There are 0 zeros 6 +0 between 5 and 5 (one is approimately 0.05). f 0 eists everywhere, so f has 0 critical numbers. 45. f() +5, [0, ]. f 0 () 6 0. Applying the Closed Interval Method, we find that f(0) 5, f() 7, and f() 4. So f(0) 5 is the absolute maimum value and f() 7 is the absolute minimum value.

4 4 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION 47. f() +, [, ]. f 0 () 6 6 6( ) 6( )( +)0,. f( ), f( ) 8, f() 9, and f() 8. So f( ) 8 is the absolute maimum value and f() 9 is the absolute minimum value. 49. f() 4 +, [, ]. f 0 () 4 4 4( ) 4( +)( ) 0, 0,. f( ), f( ), f(0), f(), f() 66. Sof() 66 is the absolute maimum value and f(±) is the absolute minimum value. 5. f(), [0, ]. + f 0 () ( +) () 0 ±,but is not in [0, ]. f(0) 0, ( +) ( +) f(), f().sof() is the absolute maimum value and f(0) 0 is the absolute minimum value f(t) t 4 t, [, ]. f 0 (t) t (4 t ) / ( t)+(4 t ) / t + 4 t t +(4 t ) 4 t. 4 t 4 t 4 t f 0 (t) 0 4 t 0 t t ±,butt is not in the given interval, [, ]. f 0 (t) does not eist if 4 t 0 t ±, but is not in the given interval. f( ), f,and f() 0. Sof is the absolute maimum value and f( ) is the absolute minimum value. 55. f(t) cost +sint, [0, π/]. f 0 (t) sint +cost sint +( sin t) ( sin t +sint ) ( sin t )(sin t +). f 0 (t) 0 sin t or sin t t π. f(0), f( π ) ,andf( π )0. So f( π ) 6 is the absolute maimum value and f( π )0is the absolute minimum value. 57. f() a ( ) b, 0, a>0, b>0. f 0 () a b( ) b ( ) + ( ) b a a a ( ) b [ b( ) + ( ) a] a ( ) b (a a b) At the endpoints, we have f(0) f() 0 [the minimum value of f ]. In the interval (0, ), f 0 () 0 a a + b. a a a f a b a a b a + b a a a a + b a + b a + b (a + b) a a + b (a + b) b b a (a + b) a a b b b (a + b). a+b a a a b b So f is the absolute maimum value. a + b a+b (a + b) 59. (a) From the graph, it appears that the absolute maimum value is about f( 0.77).9, and the absolute minimum value is about f(0.77).8.

5 SECTION 4. MAXIMUM AND MINIMUM VALUES 5 (b) f() 5 + f 0 () 5 4 (5 ). Sof 0 () 0 0, ±. 5 5 f and similarly, f (maimum) (minimum) (a) From the graph, it appears that the absolute maimum value is about f(0.75) 0., and the absolute minimum value is f(0) f() 0; that is, at both endpoints. (b) f() f 0 () + ( )+( ) 4. So f 0 () ( 4) 0 0or. 4 f(0) f() 0 (minimum), and f 4 (maimum) The density is defined as ρ mass volume 000 V (T ) (in g/cm ). But a critical point of ρ will also be a critical point of V [since dρ dt 000V dv and V is never 0],andV is easier to differentiate than ρ. dt V (T ) T T T V 0 (T ) T T. Setting this equal to 0 and using the quadratic formula to find T,weget T ± ( ).9665 Cor79.58 C. Since we are only interested in the region 0 C T 0 C, we check the density ρ at the endpoints and at.9665 C: ρ(0) ; ρ(0) ; ρ(.9665) So water has its maimum density at about.9665 C. 65. Let a , b , c , d 0.069, e , andf Then S(t) at 5 + bt 4 + ct + dt + et + f and S 0 (t) 5at 4 +4bt +ct +dt + e. We now apply the Closed Interval Method to the continuous function S on the interval 0 t 0. SinceS 0 eists for all t, the only critical numbers of S occur when S 0 (t) 0.Weusearootfinder on a CAS (or a graphing device) to find that S 0 (t) 0when t 0.855, t 4.68, t 7.9, andt ThevaluesofS at these critical numbers are S(t ) 0.9, S(t ) , S(t ) 0.47,andS(t 4) The values of S at the endpoints of the interval are S(0) 0.4 and S(0) Comparing the si numbers, we see that sugar was most epensive at t 4.68 (corresponding roughly to March 998) and cheapest at t (June 994).

6 6 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION 67. (a) v(r) k(r 0 r)r kr 0r kr v 0 (r) kr 0r kr. v 0 (r) 0 kr(r 0 r) 0 r 0or r0 (but 0 is not in the interval). Evaluating v at r0, r0,andr0,wegetv r0 8 kr 0, v r0 4 7 kr 0, and v(r 0)0.Since 4 >, v attains its maimum value at r 7 8 r0. This supports the statement in the tet. (b) From part (a), the maimum value of v is 4 7 kr 0. (c) 69. f() f 0 () for all,sof 0 () 0has no solution. Thus, f() has no critical number, so f() can have no local maimum or minimum. 7. If f has a local minimum at c,theng() f() has a local maimum at c,sog 0 (c) 0by the case of Fermat s Theorem proved in the tet. Thus, f 0 (c) g 0 (c) The Mean Value Theorem. f() 5 +, [, ]. Sincef is a polynomial, it is continuous and differentiable on R, so it is continuous on [, ] and differentiable on (, ). Alsof() 4 f(). f 0 (c) 0 + 6c 0 c, which is in the open interval (, ),soc satisfies the conclusion of Rolle s Theorem.. f(), [0, 9]. f, being the difference of a root function and a polynomial, is continuous and differentiable on [0, ), so it is continuous on [0, 9] and differentiable on (0, 9). Also,f(0) 0 f(9). f 0 (c) 0 c 0 c c conclusion of Rolle s Theorem. c 9 4, which is in the open interval (0, 9),soc 9 satisfies the 4 5. f() /. f( ) ( ) / 0f(). f 0 () /,sof 0 (c) 0has no solution. This does not contradict Rolle s Theorem, since f 0 (0) does not eist, and so f is not differentiable on (, ). 7. f(8) f(0) The values of c which satisfy f 0 (c) seem to be about c 0.8,., 4.4,and6.. 4

7 SECTION 4. THE MEAN VALUE THEOREM 7 9. (a), (b) The equation of the secant line is y ( ) y + 9. (c) f() +4/ f 0 () 4/. So f 0 (c) c 8 c,and f(c) + 4. Thus, an equation of the tangent line is y y +.. f() + +5, [, ]. f is continuous on [, ] and differentiable on (, ) since polynomials are continuous and differentiable on R. c 0, which is in (, ). f 0 (c) f(b) f(a) b a 6c + f() f( ) ( ) 0 6 6c 0. f(), [0, ]. f is continuous on R and differentiable on (, 0) (0, ),sof is continuous on [0, ] and differentiable on (0, ). f 0 f(b) f(a) (c) b a c / c / c c ± 7 f() f(0) c/ 0 ±, but only c 0 / is in (0, ). 5. f() ( ) f 0 () ( ). f(4) f() f 0 (c)(4 ) ( ) (c ) 4 6 (c ) 8 c c, which is not in the open interval (, 4). This does not (c ) contradict the Mean Value Theorem since f is not continuous at. 7. Let f() Thenf( ) 6 < 0 and f(0) > 0. Since f is a polynomial, it is continuous, so the Intermediate Value Theorem says that there is a number c between and 0 such that f(c) 0. Thus, the given equation has a real root. Suppose the equation has distinct real roots a and b with a<b.thenf(a) f(b) 0.Sincef is a polynomial, it is differentiable on (a, b) and continuous on [a, b]. By Rolle s Theorem, there is a number r in (a, b) such that f 0 (r) 0.But f 0 () for all,sof 0 () can never be 0. This contradiction shows that the equation can t have two distinct real roots. Hence, it has eactly one real root. 9. Let f() 5 + c for in [, ]. Iff has two real roots a and b in [, ],witha<b,thenf(a) f(b) 0.Since the polynomial f is continuous on [a, b] and differentiable on (a, b), Rolle s Theorem implies that there is a number r in (a, b)

8 8 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION such that f 0 (r) 0.Nowf 0 (r) r 5. Sincer is in (a, b), which is contained in [, ],wehave r <,sor < 4. It follows that r 5 < 4 5 < 0. This contradicts f 0 (r) 0, so the given equation can t have two real roots in [, ]. Hence, it has at most one real root in [, ].. (a) Suppose that a cubic polynomial P () has roots a <a <a <a 4,soP (a )P (a )P (a )P (a 4 ). By Rolle s Theorem there are numbers c, c, c with a <c <a, a <c <a and a <c <a 4 and P 0 (c )P 0 (c )P 0 (c )0. Thus, the second-degree polynomial P 0 () has three distinct real roots, which is impossible. (b) We prove by induction that a polynomial of degree n has at most n real roots. This is certainly true for n. Suppose that the result is true for all polynomials of degree n and let P () be a polynomial of degree n +. Suppose that P () has more than n +real roots, say a <a <a < <a n+ <a n+.thenp (a )P (a ) P (a n+ )0. By Rolle s Theorem there are real numbers c,...,c n+ with a <c <a,...,a n+ <c n+ <a n+ and P 0 (c ) P 0 (c n+ )0.Thus,thenth degree polynomial P 0 () has at least n +roots. This contradiction shows that P () has at most n +real roots.. By the Mean Value Theorem, f(4) f() f 0 (c)(4 ) for some c (, 4). Butforeveryc (, 4) we have f 0 (c). Putting f 0 (c) into the above equation and substituting f() 0, weget f(4) f() + f 0 (c)(4 ) 0 + f 0 (c) So the smallest possible value of f(4) is Suppose that such a function f eists. By the Mean Value Theorem there is a number 0 <c< with f 0 (c) f() f(0) 0 5. But this is impossible since f 0 () < 5 for all,sonosuchfunctioncaneist. 7. We use Eercise 6 with f() +, g() +,anda 0.Noticethatf(0) g(0) and f 0 () + < g0 () for >0.SobyEercise6,f(b) <g(b) +b<+ b for b>0. Another method: Apply the Mean Value Theorem directly to either f() + + or g() + on [0,b]. 9. Let f() sin and let b<a.thenf() is continuous on [b, a] and differentiable on (b, a). By the Mean Value Theorem, there is a number c (b, a) with sin a sin b f(a) f(b) f 0 (c)(a b) (cosc)(a b).thus, sin a sin b cos c b a a b. Ifa<b,then sin a sin b sin b sin a b a a b. Ifa b,both sides of the inequality are 0.. For >0, f() g(),sof 0 () g 0 ().For<0, f 0 () (/) 0 / and g 0 () (+/) 0 /,so again f 0 () g 0 (). However, the domain of g() is not an interval [it is (, 0) (0, )] so we cannot conclude that f g is constant (in fact it is not).

9 SECTION 4. HOW DERIVATIVES AFFECT THE SHAPE OF A GRAPH 9. Let g(t) and h(t) be the position functions of the two runners and let f(t) g(t) h(t). By hypothesis, f(0) g(0) h(0) 0 and f(b) g(b) h(b) 0,whereb is the finishing time. Then by the Mean Value Theorem, there is a time c, with0 <c<b, such that f 0 f(b) f(0) (c).butf(b) f(0) 0, sof 0 (c) 0.Since b 0 f 0 (c) g 0 (c) h 0 (c) 0,wehaveg 0 (c) h 0 (c). Soattimec, both runners have the same speed g 0 (c) h 0 (c). 4. How Derivatives Affect the Shape of a Graph. (a) f is increasing on (, ) and (4, 6). (b) f is decreasing on (0, ) and (, 4). (c) f is concave upward on (0, ). (d) f is concave downward on (, 4) and (4, 6). (e) The point of inflection is (, ).. (a) Use the Increasing/Decreasing (I/D) Test. (b) Use the Concavity Test. (c) At any value of where the concavity changes, we have an inflection point at (, f()). 5. (a) Since f 0 () > 0 on (, 5), f is increasing on this interval. Since f 0 () < 0 on (0, ) and (5, 6), f is decreasing on these intervals. (b) Since f 0 () 0at and f 0 changes from negative to positive there, f changes from decreasing to increasing and has a local minimum at.sincef 0 () 0at 5and f 0 changes from positive to negative there, f changes from increasing to decreasing and has a local maimum at There is an inflection point at because f 00 () changes from negative to positive there, and so the graph of f changes from concave downward to concave upward. There is an inflection point at 7because f 00 () changes from positive to negative there, and so the graph of f changes from concave upward to concave downward. 9. (a) f() + 6 f 0 () ( + 6) 6( +)( ). We don t need to include the 6 inthecharttodeterminethesignoff 0 (). Interval + f 0 () f < + increasing on (, ) << + decreasing on (, ) > increasing on (, ) (b) f changes from increasing to decreasing at and from decreasing to increasing at. Thus, f( ) 8 is a local maimum value and f() 44 is a local minimum value. (c) f 0 () f 00 () +6. f 00 () 0at, f 00 () > 0 >,and f 00 () < 0 <.Thus,f is concave upward on, and concave downward on,.thereisan inflection point at,f, 7.

10 0 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION. (a) f() 4 + f 0 () ( +)( ). Interval + f 0 () f < decreasing on (, ) <<0 + + increasing on (, 0) 0 << + + decreasing on (0, ) > increasing on (, ) (b) f changes from increasing to decreasing at 0and from decreasing to increasing at and.thus, f(0) is a local maimum value and f(±) are local minimum values. (c) f 00 () 4 +/ /. f 00 () > 0 < / or >/ and f 00 () < 0 / <</. Thus, f is concave upward on, / and /, and concave downward on /, /.Thereareinflection points at ± /, 9.. (a) f() sin +cos, 0 π. f 0 () cos sin 0 cos sin sin cos tan π or 5π. Thus, f 0 () > 0 cos sin >0 cos >sin 0 << π or π <<π and f 0 () < 0 cos <sin π << 5π.Sof is increasing on 0, π and 5π, π and f 4 is decreasing on π, 5π 4 4. (b) f changes from increasing to decreasing at π and from decreasing to increasing at 5π. Thus, f π is a local maimum value and f 5π 4 is a local minimum value. (c) f 00 () sin cos 0 sin cos tan π or 7π. Divide the interval 4 4 (0, π) into subintervals with these numbers as endpoints and complete a second derivative chart. There are inflection points at π 4, 0 and 7π 4, 0. Interval f 00 () sin cos Concavity 0, π f 00 π 4 < 0 downward π, 7π f 00 (π) > 0 upward 4 4 7π, π f 00 π 4 6 < 0 downward 5. f() f 0 () ( +)( +)( ). First Derivative Test: f 0 () < 0 << and f 0 () > 0 > or <. Sincef 0 changes from positive to negative at, f( ) 7 is a local maimum value; and since f 0 changes from negative to positive at, f() is a local minimum value. Second Derivative Test: f 00 () 0. f 0 () 0 ±. f 00 ( ) 0 < 0 f( ) 7 is a local maimum value. f 00 () 0 > 0 f() is a local minimum value. Preference: For this function, the two tests are equally easy.

11 SECTION 4. HOW DERIVATIVES AFFECT THE SHAPE OF A GRAPH 7. f() + f 0 () + ( ) / ( ).Notethatf is defined for 0;thatis, for. f 0 () 0. f 0 does not eist at, 4 4 but we can t have a local maimum or minimum at an endpoint. First Derivative Test: f 0 () > 0 < and f 0 () < 0 <<. Sincef 0 changes from positive to 4 4 negative at, f is a local maimum value. 4 Second Derivative Test: f 00 () ( ) / ( ) 4. f 00 4 < 0 f is a local maimum value. Preference: The First Derivative Test may be slightly easier to apply in this case. 9. (a) By the Second Derivative Test, if f 0 () 0 and f 00 () 5 < 0, f has a local maimum at. (b) If f 0 (6) 0,weknowthatf has a horizontal tangent at 6. Knowing that f 00 (6) 0 does not provide any additional information since the Second Derivative Test fails. For eample, the first and second derivatives of y ( 6) 4, y ( 6) 4,andy ( 6) all equal zero for 6,butthefirst has a local minimum at 6, the second has a local maimum at 6, and the third has an inflection point at 6.. f 0 (0) f 0 () f 0 (4) 0 horizontal tangents at 0,, 4. f 0 () > 0 if <0 or <<4 f is increasing on (, 0) and (, 4). f 0 () < 0 if 0 << or >4 f is decreasing on (0, ) and (4, ). f 00 () > 0 if << f is concave upward on (, ). f 00 () < 0 if <or > f is concave downward on (, ) and (, ). There are inflection points when and.. f 0 () > 0 if < f is increasing on (, ). f 0 () < 0 if > f is decreasing on (, ) and (, ). f 0 ( ) 0 horizontal tangent at. lim f 0 () there is a vertical asymptote or vertical tangent (cusp) at. f 00 () > 0 if 6 f is concave upward on (, ) and (, ). 5. The function must be always decreasing (since the first derivative is always negative) and concave downward (since the second derivative is always negative).

12 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION 7. (a) f is increasing where f 0 is positive, that is, on (0, ), (4, 6),and(8, ); and decreasing where f 0 is negative, that is, on (, 4) and (6, 8). (b) f has local maima where f 0 changes from positive to negative, at and at 6, and local minima where f 0 changes from negative to positive, at 4 and at 8. (c) f is concave upward (CU) where f 0 is increasing, that is, on (, 6) and (6, ), and concave downward (CD) where f 0 is decreasing, that is, on (0, ). (e) (d) There is a point of inflection where f changes from being CD to being CU, that is, at. 9. (a) f() f 0 () 6 6 6( ) 6( )( +). f 0 () > 0 < or >and f 0 () < 0 <<. Sof is increasing on (, ) and (, ), and f is decreasing on (, ). (b) Since f changes from increasing to decreasing at, f( ) 7 is a local maimum value. Since f changes from decreasing to increasing at, f() 0 is a local minimum value. (d) (c) f 00 () 6( ) f 00 () > 0 on, and f 00 () < 0 on,. So f is concave upward on, and concave downward on,.there is a change in concavity at, and we have an inflection point at,.. (a) f() + 4 f 0 () 4 4 4( )4( + )( ). f 0 () > 0 < or 0 << and f 0 () < 0 <<0 or >. Sof is increasing on (, ) and (0, ) and f is decreasing on (, 0) and (, ). (b) f changes from increasing to decreasing at and,sof( ) and f() are local maimum values. f changes from decreasing to increasing at 0,sof(0) is a local minimum value. (c) f 00 () 4 4( ). f 00 () 0 0 (d) ±/. f 00 () > 0 on /, / and f 00 () < 0 on, / and /,.Sof is concave upward on /, / and f is concave downward on, / and /,. f ±/ There are points of inflection at ±/, 9.. (a) h() ( +) 5 5 h 0 () 5( +) 4 5. h 0 () 0 5( +) 4 5 ( +) 4 ( +) +or + 0or. h 0 () > 0 < or >0 and h 0 () < 0 <<0. Soh is increasing on (, ) and (0, ) and h is decreasing on (, 0).

13 SECTION 4. HOW DERIVATIVES AFFECT THE SHAPE OF A GRAPH (b) h( ) 7 is a local maimum value and h(0) is a local minimum value. (d) (c) h 00 () 0( +) 0. h 00 () > 0 > and h 00 () < 0 <,soh is CU on (, ) and h is CD on (, ). There is a point of inflection at (,h( )) (, ). 5. (a) A() + A 0 () ( +) / ( +) The domain of A is [, ). A 0 () > 0 for > and A 0 () < 0 for <<,soa is increasing on (, ) and decreasing on (, ). (b) A( ) is a local minimum value. + ( +6) (c) A 00 + () + (d) 6( +) ( +6) + ( +4) 4( +) / 4( +) / 4( +) / A 00 () > 0 for all >,soa is concave upward on (, ). Thereisnoinflection point. 7. (a) C() / ( +4) 4/ +4 / C 0 () 4 / + 4 / 4 4( +) / ( +). C 0 () > 0 if <<0or >0and C 0 () < 0 for <,soc is increasing on (, ) and C is decreasing on (, ). (b) C( ) is a local minimum value. (d) (c) C 00 () 4 9 / 8 9 5/ 4 4( ) 9 5/ ( ) 9. 5 C 00 () < 0 for 0 <<and C 00 () > 0 for <0and >,soc is concave downward on (0, ) and concave upward on (, 0) and (, ). There are inflection points at (0, 0) and, 6 (, 7.56). 9. (a) f(θ) cosθ +cos θ, 0 θ π f 0 (θ) sinθ +cosθ ( sin θ) sinθ ( + cos θ). f 0 (θ) 0 θ 0,π,and π. f 0 (θ) > 0 π<θ<πand f 0 (θ) < 0 0 <θ<π.sof is increasing on (π, π) and f is decreasing on (0,π). (b) f(π) is a local minimum value. (c) f 0 (θ) sinθ ( + cos θ) f 00 (θ) sinθ( sin θ)+(+cosθ)( cosθ) sin θ cosθ cos θ ( cos θ) cosθ cos θ 4cos θ cosθ + ( cos θ +cosθ ) ( cos θ )(cos θ +) Since (cos θ +)< 0 [for θ 6 π], f 00 (θ) > 0 cosθ < 0 cos θ< π <θ< 5π and

14 4 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION (d) f 00 (θ) < 0 cos θ> 0 <θ< π or 5π <θ<π. Sof is CU on π, 5π and f is CD on 0, π 5π, π. There are points of inflection at π,f π π, 5 4 and 5π,f 5π 5π, 5 4. and 4. The nonnegative factors ( +) and ( 6) 4 do not affect the sign of f 0 () ( +) ( ) 5 ( 6) 4. So f 0 () > 0 ( ) 5 > 0 > 0 >. Thus,f is increasing on the interval (, ). 4. (a) From the graph, we get an estimate of f().4 as a local maimum value, and no local minimum value. f() + + f 0 () ( +) /. f 0 () 0. f() is the eact value. (b) From the graph in part (a), f increases most rapidly somewhere between and.tofind the eact value, 4 we need to find the maimum value of f 0, which we can do by finding the critical numbers of f 0. f 00 () 0 ± corresponds to the minimum value of f 0. ( +) 5/ 4 4 Themaimumvalueoff 0 occurs at f() cos + cos f 0 () sin sin f 00 () cos cos (a) From the graph of f,itseemsthatf is CD on (0, ),CUon(,.5),CDon (.5,.7),CUon(.7, 5.), and CD on (5., π). The points of inflection appear to be at (, 0.4), (.5, 0.6), (.7, 0.6),and(5., 0.4). (b) From the graph of f 00 (and zooming in near the zeros), it seems that f is CD on (0, 0.94),CU on(0.94,.57),cdon(.57,.7),cuon(.7, 5.5), and CD on (5.5, π). Refined estimates of the inflection points are (0.94, 0.44), (.57, 0.6), (.7, 0.6), and(5.5, 0.44). 47. In Maple, we define f andthenusethecommand plot(diff(diff(f,),),-..);. InMathematica,wedefine f andthenuseplot[dt[dt[f,],],{,-,}]. Weseethatf 00 > 0 for < 0.6and >0.0 [ 0.0] andf 00 < 0 for 0.6 <<0.0. Sof is CU on (, 0.6) and (0.0, ) and CD on ( 0.6, 0.0).

15 SECTION 4. HOW DERIVATIVES AFFECT THE SHAPE OF A GRAPH (a) The rate of increase of the population is initially very small, then gets larger until it reaches a maimum at about t 8hours, and decreases toward 0 as the population begins to level off. (b) The rate of increase has its maimum value at t 8hours. (c) The population function is concave upward on (0, 8) and concave downward on (8, 8). (d) At t 8, the population is about 50,sotheinflection point is about (8, 50). 5. Most students learn more in the third hour of studying than in the eighth hour, so K() K() is larger than K(8) K(7). In other words, as you begin studying for a test, the rate of knowledge gain is large and then starts to taper off, so K 0 (t) decreases and the graph of K is concave downward. 5. f() a + b + c + d f 0 () a +b + c. We are given that f() 0 and f( ),sof() a + b + c + d 0and f( ) 8a +4b c + d.alsof 0 () a +b + c 0and f 0 ( ) a 4b + c 0byFermat s Theorem. Solving these four equations, we get a, b, c 4, d 7, so the function is f() Suppose that f is differentiable on an interval I and f 0 () > 0 for all in I ecept c. Toshowthatf is increasing on I, let, be two numbers in I with <. Case < <c.letj be the interval { I <c}. By applying the Increasing/Decreasing Test to f on J,weseethatf is increasing on J,sof( ) <f( ). Case Case c< <. Apply the Increasing/Decreasing Test to f on K { I >c}. < c. Apply the proof of the Increasing/Decreasing Test, using the Mean Value Theorem (MVT) on the interval [, ] and noting that the MVT does not require f to be differentiable at the endpoints of [, ]. Case 4 c <. Same proof as in Case. Case 5 <c<.bycasesand4,f is increasing on [,c] and on [c, ],sof( ) <f(c) <f( ). In all cases, we have shown that f( ) <f( ). Since, were any numbers in I with <,wehaveshownthatf is increasing on I. 57. (a) Since f and g are positive, increasing, and CU on I with f 00 and g 00 never equal to 0,wehavef>0, f 0 0, f 00 > 0, g>0, g 0 0, g 00 > 0 on I. Then(fg) 0 f 0 g + fg 0 (fg) 00 f 00 g +f 0 g 0 + fg 00 f 00 g + fg 00 > 0 on I fg is CU on I. (b) In part (a), if f and g are both decreasing instead of increasing, then f 0 0 and g 0 0 on I, so we still have f 0 g 0 0 on I. Thus,(fg) 00 f 00 g +f 0 g 0 + fg 00 f 00 g + fg 00 > 0 on I fg is CU on I as in part (a).

16 6 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION (c) Suppose f is increasing and g is decreasing [with f and g positive and CU]. Then f 0 0 and g 0 0 on I,sof 0 g 0 0 on I and the argument in parts (a) and (b) fails. Eample. I (0, ), f(), g() /. Then(fg)(),so(fg) 0 () and (fg) 00 () > 0 on I. Thus,fg is CU on I. Eample. I (0, ), f() 4, g() /. Then(fg)() 4,so(fg) 0 () / and (fg) 00 () / < 0 on I. Thus,fg is CD on I. Eample. I (0, ), f(), g() /. Thus,(fg)(),sofg is linear on I. 59. f() tan f 0 () sec > 0 for 0 << π since sec >for 0 << π.sof is increasing on 0, π.thus,f() >f(0) 0 for 0 << π tan >0 tan >for 0 << π. 6. Let the cubic function be f() a + b + c + d f 0 () a +b + c f 00 () 6a +b. So f is CU when 6a +b>0 > b/(a),cdwhen< b/(a), and so the only point of inflection occurs when b/(a). If the graph has three -intercepts, and, then the epression for f() must factor as f() a( )( )( ). Multiplying these factors together gives us f() a[ ( + + ) +( + + ) ] Equating the coefficients of the -terms for the two forms of f gives us b a( + + ). Hence, the -coordinate of the point of inflection is b a a( + + ) a By hypothesis g f 0 is differentiable on an open interval containing c. Since(c, f(c)) is a point of inflection, the concavity changes at c,sof 00 () changes signs at c. Hence, by the First Derivative Test, f 0 has a local etremum at c. Thus, by Fermat s Theorem f 00 (c) Using the fact that,wehavethatg() g 0 () + g 00 () / < 0 for <0and g00 () > 0 for >0,so(0, 0) is an inflection point. But g 00 (0) does not eist. 67. (a) f() 4 sin f 0 () 4 cos +sin (4 )4 sin cos. g() 4 +sin 4 + f() g 0 () 8 + f 0 (). h() 4 +sin 4 + f() h 0 () 8 + f 0 (). It is given that f(0) 0,sof 0 f() f(0) 4 sin (0) lim lim 0 lim sin Since

17 SECTION 4.4 LIMITS AT INFINITY; HORIZONTAL ASYMPTOTES 7 sin and lim 0 0,weseethatf 0 (0) 0 by the Squeeze Theorem. Also, g 0 (0) 8(0) + f 0 (0) 0 and h 0 (0) 8(0) + f 0 (0) 0,so0 is a critical number of f, g,andh. For n nπ For n+ [n a nonzero integer], sin sinnπ 0and cos cosnπ,sof 0 ( n ) n < 0. n n (n +)π, sin sin(n +)π 0and cos cos(n +)π, so n+ n+ f 0 ( n+) n+ > 0. Thus,f 0 changes sign infinitely often on both sides of 0. Net, g 0 ( n )8 n + f 0 ( n )8 n n n(8 n ) < 0 for n < 8,but g 0 ( n+) 8 n+ + n+ n+(8 n+ +)> 0 for n+ > 8,sog0 changes sign infinitely often on both sides of 0. Last, h 0 ( n ) 8 n + f 0 ( n ) 8 n n n(8 n +)< 0 for n > 8 and h 0 ( n+) 8 n+ + n+ n+( 8 n+ +)> 0 for n+ < 8,soh0 changes sign infinitely often on both sides of 0. (b) f(0) 0 and since sin and hence 4 sin is both positive and negative inifinitely often on both sides of 0,and arbitrarily close to 0, f has neither a local maimum nor a local minimum at 0. Since +sin, g() 4 +sin > 0 for 6 0,sog(0) 0 is a local minimum. Since +sin, h() 4 +sin < 0 for 6 0,soh(0) 0 is a local maimum. 4.4 Limits at Infinity; Horizontal Asymptotes. (a) As becomes large, the values of f() approach 5. (b) As becomes large negative, the values of f() approach.. (a) lim f() (b) lim f() (c) lim f() + (d) lim f() (e) lim f() (f ) Vertical:, ; Horizontal: y, y 5. If f() /, then a calculator gives f(0) 0, f() 0.5, f(), f().5, f(4), f(5) 0.785, f(6) 0.565, f(7) 0.885, f(8) 0.5, f(9) , f(0) , f(0) , f(50).04 0, f(00) It appears that lim / 0.

18 8 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION 7. lim lim ( +4)/ [divide both the numerator and denominator by ( +5 8)/ (the highest power of that appears in the denominator)] lim ( / +4/ ) lim ( + 5/ 8/ ) lim lim (/)+ lim (4/ ) lim + lim (5/) lim (8/ ) lim (/)+4 lim (/ ) +5 lim(/) 8 lim (/ ) 0 + 4(0) +5(0) 8(0) [Limit Law 5] [Limit Laws and ] [Limit Laws 7 and ] [Theorem 4] 9. lim + lim / lim (/) ( +)/ lim ( + /). lim 7 lim (/) lim + lim (/) 0 ( )/ lim (/ / ) lim ( 7)/ lim ( 7/ ) lim (/ ) lim (/) lim lim 7 lim (/ ) 0 0 7(0) +(0) 0 0. Divide both the numerator and denominator by (the highest power of that occurs in the denominator). lim lim lim +5 lim lim lim lim +4 lim lim lim +5(0) 0 + 4(0) 5. First, multiply the factors in the denominator. Then divide both the numerator and denominator by u 4. lim u 4u 4 +5 (u )(u ) lim u lim u 4u 4 +5 u 4 5u + lim u lim 4+ 5u 4 5u + u 4 u 4u 4 +5 u 4 u 4 5u + u 4 lim u lim 4+5 lim u u lim 5 lim u u u 4 + lim u u 4+ 5 u 4 5 u + u 4 u 4 4+5(0) 5(0) + (0) 4

19 SECTION 4.4 LIMITS AT INFINITY; HORIZONTAL ASYMPTOTES / lim (96 )/ 6 7. lim lim + ( +)/ lim lim 9 / 5 lim lim (/ ) ( + / ) lim 9 lim (/5 ) +0 [since 6 for >0] 9. lim 9 + lim lim lim. lim + a + b lim lim / 9 / + / +/ lim lim lim lim / / 9+/ () a + b + a + + b + a + + b ( + a) ( + b) + a + + b lim a b +a/ + +b/ [(a b)]/ + a + + b / a b a b lim + lim ( )/ 4 [divide by the highest power of in the denominator] 4 ( + 4 )/ 4 lim / +/ + / 4 / + because (/ +/ + ) and (/ 4 / +) as. 5. lim (4 + 5 ) lim 5 ( +) [factor out the largest power of ] because 5 and / + as. Or: lim lim 4 ( + ). 7. lim lim since and as. 9. If t,then lim sin lim sin t sin t lim. t 0 + t t 0 + t. (a) (b) f() 0, , ,000, From the graph of f() + ++,weestimate the value of lim f() to be 0.5. Fromthetable,weestimatethelimit to be 0.5.

20 0 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION (c) lim + ++ lim lim lim + + ( +)(/) + + (/) lim +(/) (/)+(/ ) Note that for <0, wehave, so when we divide the radical by, with<0, weget (/)+(/ ). +. lim lim + + lim lim + lim lim + lim lim lim +0, soy is a horizontal asymptote. 0 The denominator is zero when and the numerator is not zero, so we investigate y f() + as approaches. lim f() because as the numerator is positive and the denominator approaches 0 through negative values. Similarly, lim f(). Thus, is a vertical asymptote. + The graph confirms our work lim + lim + y f() + + lim lim + lim lim + lim + lim + lim lim ( )( +) ( +)( ) lim + + lim +0 0, soy is a horizontal asymptote. +0 (0),so lim f(), f(), lim f(),and lim f(). Thus, + + and are vertical asymptotes. The graph confirms our work. 7. y f() 6 +5 ( ) ( +)( ) ( +) g() for 6. ( )( 5) ( )( 5) 5 The graph of g is the same as the graph of f with the eception of a hole in the graph of f at.bylongdivision,g() As ±, g() ±, so there is no horizontal asymptote. The denominator of g is zero when 5. lim g() and lim g(),so 5is a vertical asymptote. The graph confirms our work.

21 SECTION 4.4 LIMITS AT INFINITY; HORIZONTAL ASYMPTOTES 9. From the graph, it appears y is a horizontal asymptote. lim lim lim +0, soy isahorizontal asymptote The discrepancy can be eplained by the choice of the viewing window. Try [ 00,000, 00,000] by [, 4] to get a graph that lends credibility to our calculation that y is a horizontal asymptote. +(500/) +(500/) + (00/ )+(000/ ) 4. Let s look for a rational function. () lim f() 0 ± degree of numerator < degree of denominator () lim f() 0 there is a factor of in the denominator (not just, since that would produce a sign change at 0), and the function is negative near 0. () lim f() and lim f() + vertical asymptote at ;thereisafactorof( ) in the denominator. (4) f() 0 is an -intercept; there is at least one factor of ( ) in the numerator. Combining all of this information and putting in a negative sign to give us the desired left- and right-hand limits gives us f() as one possibility. ( ) 4. y has domain (, ) (, ). + lim ± + lim / ± / + 0,soy is a HA. 0+ The line is a VA. y 0 ( + )( ) ( )() < 0 for 6. Thus, ( + ) ( + ) (, ) and (, ) are intervals of decrease. y 00 ( + ) [( + ) ] 4 ( + ) < 0 for < and y00 > 0 for >,sothecurveiscdon(, ) and CU on (, ). Since is not in the domain, there is no IP. 45. lim ± + lim horizontal asymptote. ± / +/ 0 0,soy 0is a +0 y 0 + () ( +) ( +) 0when ± and y0 > 0 < <<,soy is increasing on (, ) and decreasing on (, ) and (, ). [continued]

22 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION y 00 ( + ) ( ) ( )( +) ( ) ( + ) 4 ( + ) > 0 > or <<0, soy is CU on, and, 0 and CD on, and 0,. 47. y f() ( ) 4 ( + )( ). They-intercept is f(0) 0. The-intercepts are 0,,and [found by solving f() 0for ]. Since 4 > 0 for 6 0, f doesn t change sign at 0. The function does change sign at and.as ±, f() 4 ( ) approaches because 4 and ( ). 49. y f() ( )( + ) ( ) 4.They-intercept is f(0) () () 4. The -intercepts are,,and. There is a sign change at,butnotat and. When is large positive, is negative and the other factors are positive, so lim f(). When is large negative, is positive, so lim f(). 5. First we plot the points which are known to be on the graph: (, ) and (0, 0). We can also draw a short line segment of slope 0 at,since we are given that f 0 () 0. Now we know that f 0 () < 0 (that is, the function is decreasing) on (0, ),andthatf 00 () < 0 on (0, ) and f 00 () > 0 on (, ).Sowemustjointhepoints(0, 0) and (, ) in such a way that the curve is concave down on (0, ) and concave up on (, ). The curve must be concave up and increasing on (, 4) and concave down and increasing toward y on (4, ). Now we just need to reflect the curve in the y-ais, since we are given that f is an even function [the condition that f( ) f() for all ]. 5. We are given that f() f 0 () 0. So we can draw a short horizontal line at the point (, 0) to represent this situation. We are given that 0and are vertical asymptotes, with lim f(), lim f() and lim f(),so 0 + we can draw the parts of the curve which approach these asymptotes. On the interval (, 0), the graph is concave down, and f() as. Between the asymptotes the graph is concave down. On the interval (, ) the graph is concave up, and f() 0 as,so y 0is a horizontal asymptote. The diagram shows one possible function satisfying all of the given conditions.

23 SECTION 4.4 LIMITS AT INFINITY; HORIZONTAL ASYMPTOTES 55. (a) Since sin for all, sin for >0. As, / 0 and / 0, so by the Squeeze sin Theorem, (sin )/ 0. Thus, lim 0. (b) From part (a), the horizontal asymptote is y 0. The function y (sin)/ crosses the horizontal asymptote whenever sin 0; that is, at πn for every integer n. Thus, the graph crosses the asymptote an infinite number of times. 57. Divide the numerator and the denominator by the highest power of in Q(). (a) If deg P<deg Q, then the numerator 0 but the denominator doesn t. So lim [P ()/Q()] 0. (b) If deg P>deg Q, then the numerator ± but the denominator doesn t, so lim [P ()/Q()] ± (depending on the ratio of the leading coefficients of P and Q) lim lim and lim lim Therefore, by the Squeeze Theorem, lim f() Let g() + and f() g().5. Note that + + lim g() and lim f() 0. We are interested in finding the -value at which f() < From the graph, we find that 4.804, so we choose N 5(or any larger number). 6. For ε 0.5,weneedtofind N such that ( ) < < <.5 whenever N. We graph the three parts of this + inequality on the same screen, and see that the inequality holds for 6. So we choose N 6 (or any smaller number). 4 + For ε 0.,weneed. < <.9 whenever N. From the + graph, it seems that this inequality holds for. So we choose N (oranysmallernumber). 65. (a) / < > / >00 ( >0) (b) If ε>0is given, then / <ε > /ε >/ ε.letn / ε. Then >N > ε 0 <ε,so lim 0.

24 4 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION 67. For <0, / 0 /. Ifε>0 is given, then / < ε < /ε. Take N /ε. Then<N < /ε (/) 0 / < ε,so lim (/) Suppose that lim f() L. Then for every ε>0there is a corresponding positive number N such that f() L <ε whenever >N.Ift /,then>n 0 < / < /N 0 <t</n. Thus, for every ε>0there is a corresponding δ>0 (namely /N ) such that f(/t) L <εwhenever 0 <t<δ.thisprovesthat lim f(/t) L lim f(). t 0 + Now suppose that lim f() L. Then for every ε>0there is a corresponding negative number N such that f() L <εwhenever <N.Ift /,then<n /N < / < 0 /N < t < 0. Thus, for every ε>0 there is a corresponding δ>0 (namely /N ) such that f(/t) L <εwhenever δ <t<0.thisprovesthat lim t 0 f(/t) L lim f(). 4.5 Summary of Curve Sketching. y f() + ( +) A. f is a polynomial, so D R. H. B. -intercept 0, y-intercept f(0) 0 C. f( ) f(),so f is odd; the curve is symmetric about the origin. D. f is a polynomial, so there is no asymptote. E. f 0 () +> 0,sof is increasing on (, ). F. There is no critical number and hence, no local maimum or minimum value. G. f 00 () 6>0on (0, ) and f 00 () < 0 on (, 0),sof is CU on (0, ) and CD on (, 0). Since the concavity changes at 0,thereisan inflection point at (0, 0).. y f() 5 +9 ( ) 7 + A. D R B. y-intercept: f(0) ; -intercepts: f() 0 or (by the quadratic formula) 7 ± , 6.85 C. No symmetry D. No asymptote E. f 0 () ( 6 +5) ( )( 5) > 0 <<5 H. so f is increasing on (, 5) and decreasing on (, ) and (5, ). F. Local maimum value f(5) 7, local minimum value f() 5 G. f 00 () 8 6 6( ) > 0 <,sof is CU on (, ) and CD on (, ). IPat(, )

25 SECTION 4.5 SUMMARY OF CURVE SKETCHING 5 5. y f() 4 +4 ( +4) A. D R B. y-intercept: f(0) 0; H. -intercepts: f() 0 4, 0 C. No symmetry D. No asymptote E. f 0 () ( +)> 0 >,sof is increasing on (, ) and decreasing on (, ). F. Local minimum value f( ) 7, no local maimum G. f 00 () +4 ( +)< 0 <<0,sof is CD on (, 0) and CU on (, ) and (0, ). IPat(0, 0) and (, 6) 7. y f() A. D R B. y-intercept: f(0) C. No symmetry D. No asymptote E. f 0 () ( ) 0( )( + +),sof 0 () < 0 0 << and f 0 () > 0 <0 or >. Thus,f is increasing on (, 0) and (, ) and decreasing on (0, ). F. Local maimum value f(0), local minimum value f() G. f 00 () (4 ) H. so f 00 () 0 / 4. f 00 () > 0 >/ 4 and f 00 () < 0 </ 4,sof is CD on, / 4 and CU on / 4, 9.IPat, 4 4 (0.60, 0.786) 9. y f() /( ) A. D { 6 } (, ) (, ) B. -intercept 0, y-intercept f(0) 0 C. No symmetry D. lim ±,soy is a HA. lim, lim +,so is a VA. E. f 0 () ( ) < 0 for 6,sof is ( ) ( ) H. decreasing on (, ) and (, ). F. No etreme values G. f 00 () > 0 ( ) >,sof is CU on (, ) and CD on (, ). NoIP. y f() /( 9) A. D { 6 ±} (, ) (, ) (, ) B. y-intercept f(0) 9,no -intercept C. f( ) f() f is even; the curve is symmetric about the y-ais. D. lim 0,soy 0 ± 9 is a HA. lim, lim 9 + 9, lim 9, lim, so and + 9 are VA. E. f 0 () ( > 0 <0 ( 6 )sof is increasing on (, ) and (, 0) and 9)

26 6 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION decreasing on (0, ) and (, ). F. Local maimum value f(0) 9. H. G. y 00 ( 9) +()( 9)() ( 9) 4 6( +) ( 9) > 0 > 9 > or <,sof is CU on (, ) and (, ) and CD on (, ). NoIP. y f() /( +9) A. D R B. y-intercept: f(0) 0; -intercept: f() 0 0 C. f( ) f(),sof is odd and the curve is symmetric about the origin. D. lim ± [/( +9)]0,soy 0is a HA; no VA E. f 0 () ( +9)() () 9 ( + )( ) > 0 <<,sof is increasing ( +9) ( +9) ( +9) on (, ) and decreasing on (, ) and (, ). F. Local minimum value f( ) 6, local maimum value f() 6 f 00 () ( +9) ( ) (9 ) ( +9)() ()( +9)[ ( +9) (9 )] ( 7) [( +9) ] ( +9) 4 ( +9) 0 0, ± 7 ± G. f 00 () ( +9) ( ) (9 ) ( + 9)() ()( +9) ( +9) (9 ) [( +9) ] ( +9) 4 ( 7) ( +9) 0 0, ± 7 ± H. f 00 () > 0 <<0 or >,sof is CU on, 0 and,, and CD on, and 0,. There are three inflection points: (0, 0) and ±, ±. 5. y f() A. D { 6 0} (, 0) (0, ) B. No y-intercept; -intercept: f() 0 C. No symmetry D. lim 0,soy 0is a HA. lim,so 0is a VA. ± 0 E. f 0 () ( ) + ( ),sof 0 () > 0 0 <<and f 0 () < 0 ( ) 4 <0 or >. Thus,f is increasing on (0, ) and decreasing on (, 0) H. and (, ). F. No local minimum, local maimum value f() 4. G. f 00 () ( ) [ ( )] 6 ( ). ( ) 6 4 f 00 () is negative on (, 0) and (0, ) and positive on (, ),sof is CD on (, 0) and (0, ) and CU on (, ). IPat, 9

27 SECTION 4.5 SUMMARY OF CURVE SKETCHING 7 7. y f() + ( +) A. D R B. y-intercept: f(0) 0; + + -intercepts: f() 0 0 C. f( ) f(),sof is even; the graph is symmetric about the y-ais. D. lim ± +,soy is a HA. No VA. E. Using the Reciprocal Rule, f 0 () ( +) 6 ( +). f 0 () > 0 >0and f 0 () < 0 <0,sof is decreasing on (, 0) and increasing on (0, ). F. Local minimum value f(0) 0, no local maimum. G. f 00 () ( +) 6 6 ( +) [( +) ] H. 6( +)[( +) 4 ] 6( ) 8( +)( ) ( +) 4 ( +) ( +) f 00 () is negative on (, ) and (, ) and positive on (, ), so f is CD on (, ) and (, ) and CU on (, ). IPat ±, 4 9. y f() 5 A. The domain is { 5 0} (, 5] B. y-intercept: f(0) 0; -intercepts: f() 0 0, 5 C. No symmetry D. No asymptote E. f 0 () (5 ) / ( ) + (5 ) / (5 ) / [ +(5 )] < 0,sof is increasing on, 0 and decreasing on 0 F. Local maimum value f 0, ; no local minimum G. f 00 () (5 )/ ( ) (0 ) (5 ) / ( ) 5 (5 ) / [ 6(5 )+(0 )] 4(5 ) f 00 () < 0 for <5,sof is CD on (, 5). NoIP 0 4(5 ) / H. 0 5 > 0. y f() + ( +)( ) A. D { ( +)( ) 0} (, ] [, ) B. y-intercept: none; -intercepts: and C. No symmetry D. No asymptote E. f 0 () ( + ) / + ( +) +, f 0 () 0if,but is not in the domain. f 0 () > 0 > and f 0 () < 0 <, so (considering the domain) f is increasing on (, ) and f is decreasing on (, ). F. No local etrema G. f 00 () ( + ) / () ( +) ( + ) / ( +) + H. ( + ) / 4( + ) (4 +4 +) 4( + ) 9 4( + ) / < 0 so f is CD on (, ) and (, ). NoIP

28 8 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION. y f() / + A. D R B. y-intercept: f(0) 0; -intercepts: f() 0 0 C. f( ) f(),sof is odd; the graph is symmetric about the origin. D. lim f() lim and lim f() lim + lim + lim / +/ lim / +/ so y ± are HA. +0 lim / +/ lim / +/ lim +/ +0 +/ No VA. + E. f 0 () + + [( +) / ] ( +) > 0 for all,sof is increasing on R. / ( / +) F. No etreme values H. G. f 00 () ( +) 5/ ( +),sof 00 () > 0 for <0 5/ and f 00 () < 0 for >0. Thus,f is CU on (, 0) and CD on (0, ). IP at (0, 0) 5. y f() / A. D {, 6 0} [, 0) (0, ] B. -intercepts ±, noy-intercept C. f( ) f(), so the curve is symmetric about (0, 0). D. lim 0 + so 0is a VA. E. f 0 / () on (, 0) and (0, ). F. No etreme values G. f 00 () ( ) f is CU on, IP at ±, ± > 0 << / and 0, and CD on, 0 and or 0 <<,so,., lim 0 < 0,sof is decreasing H., 7. y f() / A. D R B. y-intercept: f(0) 0; -intercepts: f() 0 / ( 7) 0 0, ± C. f( ) f(),sof is odd; the graph is symmetric about the origin. D. No asymptote E. f 0 () / /. / / f 0 () > 0 when > and f 0 () < 0 when 0 < <,sof is increasing on (, ) and (, ),and decreasing on (, 0) and (0, ) [hence decreasing on (, ) since f is H. continuous on (, )]. F. Local maimum value f( ), local minimum value f() G. f 00 () 5/ < 0 when <0 and f 00 () > 0 when >0,sof is CD on (, 0) and CU on (0, ). IPat(0, 0)

29 SECTION 4.5 SUMMARY OF CURVE SKETCHING 9 9. y f() A. D R B. y-intercept: f(0) ; -intercepts: f() 0 0 ± C. f( ) f(), so the curve is symmetric about the y-ais. D. No asymptote E. f 0 () ( ) / () ( ). f 0 () > 0 >0and f 0 () < 0 <0,sof is increasing on (0, ) and decreasing on (, 0). F. Local minimum value f(0) G. f 00 () ( ) / () ( ) / () H. [( ) / ] 9 ( ) / [( ) 4 ] ( ) 4/ ( +) 9( ) 5/ f 00 () > 0 << and f 00 () < 0 < or >,so f is CU on (, ) and f is CD on (, ) and (, ). IPat(±, 0). y f() sin sin A. D R B. y-intercept: f(0) 0; -intercepts: f() 0 sin ( sin )0 sin 0 [since sin < ] nπ, n an integer. C. f( ) f(),sof is odd; the graph (shown for π π) is symmetric about the origin and periodic with period π. D. No asymptote E. f 0 () cos sin cos cos ( sin )cos. f 0 () > 0 cos >0 nπ π, nπ + π for each integer n, andf 0 () < 0 cos <0 nπ + π, nπ + π for each integer n. Thus, f is increasing on nπ π, nπ + π for each integer n, and f is decreasing on nπ + π, nπ + π for each integer n. F. f has local maimum values f(nπ + π )and local minimum values f(nπ + π ). G. f 00 () 9sin cos 9sin ( sin ) 9sin ( sin )( + sin ). f 00 () < 0 sin >0and sin 6 ± nπ, nπ + π nπ + π, nπ + π for some integer n. f 00 () > 0 sin <0and sin 6 ± (n )π, (n )π + π (n )π + π, nπ for some integer n. Thus, f is CD on the intervals nπ, n + π and H. n + π, (n +)π [hence CD on the intervals (nπ, (n +)π)] for each integer n,andf is CU on the intervals (n )π, n π and n π, nπ [hence CU on the intervals ((n )π, nπ)] for each integer n. f has inflection points at (nπ, 0) for each integer n.. y f() tan, π << π A. D π, π B. Intercepts are 0 C. f( ) f(),sothecurveis symmetric about the y-ais. D. lim tan and lim tan,so π and π are VA. (π/) (π/) + E. f 0 () tan + sec >0 0 << π,sof increases on 0, π H. and decreases on π, 0. F. Absolute and local minimum value f(0) 0. G. y 00 sec +tan sec >0for π << π,sofis CU on π, π.noip

30 40 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION 5. y f() sin, 0 <<π A. D (0, π) B. No y-intercept. The -intercept, approimately.9, can be found using Newton s Method. C. No symmetry D. No asymptote E. f 0 () cos >0 cos < π << 5π or 7π <<π, sof is increasing on π, 5π and 7π, π and decreasing on 0, π and 5π, 7π. F. Local minimum value f π π 6, local maimum value H. f 5π 5π +, local minimum value f 7π 6 7π 6 G. f 00 () sin>0 0 <<πor π <<π,sof is CU on (0,π) and (π, π) and CD on (π, π). IPsat π, π and (π, π) 7. y f() sin +cos when cos 6 sin +cos cos sin ( cos ) cos sin cos sin csc cot A. The domain of f is the set of all real numbers ecept odd integer multiples of π. B. y-intercept: f(0) 0; -intercepts: nπ, n an even integer. C. f( ) f(),sof is an odd function; the graph is symmetric about the origin and has period π. D. When n is an odd integer, lim f() and lim (nπ) integer n. No HA. E. f 0 () (nπ) + f(),so nπ is a VA for each odd ( + cos ) cos sin ( sin ) ( + cos ) +cos ( + cos ) +cos. f 0 () > 0 for all ecept odd multiples of π,sof is increasing on ((k )π, (k +)π) for each integer k. F. No etreme values G. f 00 () sin > 0 sin >0 ( + cos ) H. (kπ, (k +)π) and f 00 () < 0 on ((k )π, kπ) for each integer k. f is CU on (kπ, (k +)π) and CD on ((k )π, kπ) for each integer k. f has IPs at (kπ, 0) for each integer k. 9. m f(v) m 0 v /c.them-intercept is f(0) m 0.Therearenov-intercepts. lim f(v),sov c is a VA. v c f 0 (v) m 0( v /c ) / ( v/c ) increasing on (0,c). There are no local etreme values. m 0v c ( v /c ) / f 00 (v) (c v ) / (m 0c) m 0cv (c v ) / ( v) [(c v ) / ] m 0c(c v ) / [(c v )+v ] (c v ) m 0c(c +v ) (c v ) 5/ > 0, so f is CU on (0,c). Therearenoinflection points. m 0v c (c v ) / c m 0cv > 0,sof is (c v ) /

31 4. y W 4EI 4 + WL EI WL 4EI W 4EI L + L W 4EI ( L) c ( L) where c W isanegativeconstantand0 L. Wesketch 4EI f() c ( L) for c. f(0) f(l) 0. SECTION 4.5 SUMMARY OF CURVE SKETCHING 4 f 0 () c [( L)] + ( L) (c) c( L)[ +( L)] c( L)( L). Sofor0 <<L, f 0 () > 0 ( L)( L) < 0 [since c<0] L/ <<Land f 0 () < 0 0 <<L/. Thus, f is increasing on (L/,L) and decreasing on (0,L/), and there is a local and absolute minimum at the point (L/,f(L/)) L/,cL 4 /6. f 0 () c[( L)( L)] f 00 () c[( L)( L)+()( L)+( L)()] c(6 6L + L )0 6L ± L L ± 6 L, and these are the -coordinates of the two inflection points. 4. y +. Long division gives us: Thus, y f() and f() ( ) + + So the line y is a slant asymptote (SA). + [for 6 0] 0 as ±. 45. y Long division gives us: Thus, y f() and f() ( ) [for 6 0] 0 as ±. So the line y is a SA. 47. y f() A. D R 6,, B. y-intercept: f(0) ; -intercepts: f() ± 7 4 C. No symmetry D. lim f() and lim f(),so is a VA. (/) (/) + [continued] 0.,.8.

32 4 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION lim ± [f() ( +)] lim E. f 0 () 0,sotheliney +is a SA. ± ( ) < 0 for 6,sof is decreasing on, H. and,. F. No etreme values G. f 0 () ( ) f 00 () ( )( ) 8 () ( ),sof 00 () > 0 when > and f 00 () < 0 when <. Thus, f is CU on, and CD on,.noip 49. y f() ( +4)/ +4/ A. D { 6 0} (, 0) (0, ) B. No intercept C. f( ) f() symmetry about the origin D. lim ( +4/) but f() 4/ 0 as ±, so y is a slant asymptote. lim ( +4/) and H. 0 + lim ( +4/),so 0 is a VA. E. f 0 () 4/ > 0 0 > 4 >or <,sof is increasing on (, ) and (, ) and decreasing on (, 0) and (0, ). F. Local maimum value f( ) 4, local minimum value f() 4 G. f 00 () 8/ > 0 >0 so f is CU on (0, ) and CD on (, 0). NoIP 5. y f() A. D R B. y-intercept: f(0) ; -intercept: f() ( + )( +) C. No symmetry D. No VA lim ± [f() ( + )] lim ± + lim ± / 0, so the line y +is a slant asymptote. +/ E. f 0 () + ( +)( ) ( )() (4 + +) ( +) ( +) ( +) ( +) ( +) so f 0 () > 0 if 6 0.Thus,f is increasing on (, 0) and (0, ). Sincef is continuous at 0, f is increasing on R. F. No etreme values G. f 00 () ( +) (8 +) ( 4 +6 ) ( +)() [( +) ] 4( +)[( + )( +) 4 6 ] 4( +) ( +) 4 ( +) so f 00 () > 0 for < and 0 <<,andf 00 () < 0 for H. <<0 and >. f is CU on, and 0,, and CD on, 0 and,. There are three IPs: (0, ),, + (.7,.60),and, + (.7,.60).

33 SECTION 4.5 SUMMARY OF CURVE SKETCHING 4 5. y f() 4 +9 f 0 () f / 4 () +9 4(4 +9) 6 6. f is defined on (, ) (4 +9) / (4 +9) / f( ) f(),sof is even, which means its graph is symmetric about the y-ais. The y-intercept is f(0). Thereareno -intercepts since f() > 0 for all. lim lim (4 +9) 4 lim lim and, similarly, lim lim , so y ± are slant asymptotes. f is decreasing on (, 0) andincreasingon(0, ) with local minimum f(0). f 00 () > 0 for all,sof is CU on R. a y lim b y ± b a a.now b a a b a b a lim a a + a + b a lim which shows that y b is a slant asymptote. Similarly, a lim b a a ba b a lim 57. lim ± f() lim ± 4 + a a + 0, a a + 0,soy b is a slant asymptote. a 4 lim ± 0,sothegraphoff is asymptotic to that of y. A. D { 6 0} B. No intercept C. f is symmetric about the origin. D. lim 0 + and lim +,so 0is a vertical asymptote, and as shown above, the graph of f is asymptotic to that of y. 0 + E. f 0 () / > 0 4 > >,sof is increasing on 4, and 4 4, and decreasing on 4, 0 and 0, 4. F. Local maimum value f 4 5/4, local minimum value f /4 H. G. f 00 () 6 +/ > 0 >0,sof is CU on (0, ) and CD on (, 0). NoIP

34 44 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION 4.6 Graphing with Calculus and Calculators. f() f 0 () f 00 () f() 0 0.5,.60; f 0 () 0 0.9,.5,.58 and f 00 () 0.46,.54. From the graphs of f 0,weestimatethatf 0 < 0 and that f is decreasing on (, 0.9) and (.5,.58),andthatf 0 > 0 and f is increasing on (0.9,.5) and (.58, ) with local minimum values f(0.9) 5. and f(.58).998 and local maimum value f(.5) 4. The graphs of f 0 make it clear that f has a maimum and a minimum near.5,shownmore clearly in the fourth graph. From the graph of f 00,weestimatethatf 00 > 0 and that f is CU on (,.46) and (.54, ),andthatf 00 < 0 and f is CD on (.46,.54). There are inflection points at about (.46,.40) and (.54,.999).. f() f 0 () f 00 ()

35 SECTION 4.6 GRAPHING WITH CALCULUS AND CALCULATORS 45 From the graph of f 0,weestimatethatf is decreasing on (, 5), increasing on ( 5, 4.40), decreasing on (4.40, 8.9), and increasing on (8.9, ), with local minimum values of f( 5) 9,700,000 and f(8.9),700,000 and local maimum value f(4.40) 5,800. Fromthegraphoff 00,weestimatethatf is CU on (,.4), CDon(.4, 0), CUon(0,.9), CDon(.9, 5.08), andcuon(5.08, ). There is an inflection point at (0, 0) and at about (.4, 6,50,000), (.9,,800),and(5.08, 8,50,000). 5. f() 4 + f 0 () + + f 00 () ( ) ( 4 +) ( 4 +) We estimate from the graph of f that y 0is a horizontal asymptote, and that there are vertical asymptotes at.7, 0.4,and.46. Fromthegraphoff 0, we estimate that f is increasing on (,.7), (.7, 0.4),and(0.4, ), and that f is decreasing on (,.46) and (.46, ). There is a local maimum value at f(). From the graph of f 00,we estimate that f is CU on (,.7), ( 0.506, 0.4),and(.46, ),andthatf is CD on (.7, 0.506) and (0.4,.46). There is an inflection point at ( 0.506, 0.9). 7. f() 4 +7cos, 4 4. f 0 () 4 7sin f 00 () 7cos. f() 0.0; f 0 () 0.49,.07,or.89; f 00 () 0 ± cos 7 ±.8. From the graphs of f 0,weestimatethatf is decreasing (f 0 < 0) on ( 4,.49), increasing on (.49,.07), decreasing on (.07,.89), and increasing on (.89, 4), with local minimum values f(.49) 8.75 and f(.89) 9.99 and local maimum value f(.07) 8.79 (notice the second graph of f). From the graph of f 00,weestimatethatf is CU (f 00 > 0) on ( 4,.8), CDon(.8,.8), and CU on (.8, 4). Thereareinflection points at about (.8, 8.77) and (.8,.48).

36 46 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION 9. f() f 0 () ( +6 +) f 00 () ( +4 +6). From the graphs, it appears that f increases on ( 5.8, 0.) and decreases on (, 5.8), ( 0., 0),and(0, );thatf has a local minimum value of f( 5.8) 0.97 and a local maimum value of f( 0.) 7;thatf is CD on (, 4) and ( 0.5, 0) and is CU on ( 4, 0.5) and (0, );andthatf has IPs at ( 4, 0.97) and ( 0.5, 60). To find the eact values, note that f ± 56 8 ± 6 [ 0.9 and 5.8]. f 0 is positive (f is increasing) on 8 6, 8+ 6 and f 0 is negative (f is decreasing) on, 8 6, 8+ 6, 0,and(0, ). f ± ± 8 [ 0.5 and.75]. f 00 is positive (f is CU) on 8, + 8 and (0, ) and f 00 is negative (f is CD) on, 8 and + 8, 0.. f() lim ( +4)( ) 4 ( ) f() and lim f(). + f() +4 4 ( ) ( ) as ±,sof is asymptotic to the -ais. has VA at 0and at since lim 0 f(), dividing numerator and denominator by ( + 4/)( /) ( ) Since f is undefined at 0,ithasnoy-intercept. f() 0 ( +4)( ) 0 4 or,sof has -intercepts 4 and. Note, however, that the graph of f is only tangent to the -ais and does not cross it at,sincef is positive as and as +. 0 From these graphs, it appears that f has three maimum values and one minimum value. The maimum values are

37 SECTION 4.6 GRAPHING WITH CALCULUS AND CALCULATORS 47 approimately f( 5.6) 0.08, f(0.8) 8.5 and f(5.) and we know (since the graph is tangent to the -ais at ) that the minimum value is f() 0.. f() ( +) ( ) ( 4) 4 f 0 () ( +) ( ) ( ) ( 4) 5 [from CAS]. From the graphs of f 0, it seems that the critical points which indicate etrema occur at 0, 0.,and.5, as estimated in Eample. (There is another critical point at,butthesignoff 0 does not change there.) We differentiate again, obtaining f 00 () ( +)( ) ( ) 4 ( 4) 6. From the graphs of f 00, it appears that f is CU on ( 5., 5.0), (, 0.5), ( 0., ), (, 4) and (4, ) and CD on (, 5.), ( 5.0, ) and ( 0.5, 0.). We check back on the graphs of f to find the y-coordinates of the inflection points, and find that these points are approimately ( 5., 0.05), ( 5.0, 0.005), (, 0), ( 0.5, ), and ( 0., ). 5. y f() + +. FromaCAS,y0 + ( + +) and y / ( + +). f 0 () 0 0.4,sof is increasing on (0, 0.4) and decreasing on (0.4, ). There is a local maimum value of f(0.4) 0.4. f 00 () ,sof is CD on (0, 0.94) and CU on (0.94, ). Thereisaninflection point at (0.94, 0.4).

38 48 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION 7. y f() +5sin, 0. From a CAS, y 0 5cos + +5sin and y00 0 cos +5sin +0 sin +6 4( +5sin) /. We ll start with a graph of g() +5sin. Notethatf() g() is only defined if g() 0. g() 0 0 or 4.9, 4.0, 4.0, and 4.9. Thus, the domain of f is [ 4.9, 4.0] [0, 4.0] [4.9, 0]. From the epression for y 0, we see that y 0 0 5cos +0 cos 5.77 and π 4.5 (not in the domain of f ). The leftmost zero of f 0 is π 4.5. Moving to the right, the zeros of f 0 are, +π, +π, +4π, and +4π. Thus, f is increasing on ( 4.9, 4.5), decreasing on ( 4.5, 4.0),increasingon(0,.77), decreasing on (.77, 4.0),increasingon(4.9, 8.06), decreasing on (8.06, 0.79), increasing on (0.79, 4.4), decreasing on (4.4, 7.08), and increasing on (7.08, 0). The local maimum values are f( 4.5) 0.6, f(.77).58, f(8.06).60, andf(4.4) 4.9. The local minimum values are f(0.79).4 and f(7.08).49. f is CD on ( 4.9, 4.0), (0, 4.0), (4.9, 9.60),CUon(9.60,.5),CD on (.5, 5.8),CUon(5.8, 8.65), and CD on (8.65, 0). Thereare inflection points at (9.60,.95), (.5,.7), (5.8,.9),and(8.65, 4.0). 9. From the graph of f() sin( +sin) in the viewing rectangle [0,π] by [.,.], it looks like f has two maima and two minima. If we calculate and graph f 0 () [cos( +sin)] ( + cos ) on [0, π], we see that the graph of f 0 appears to be almost tangent to the -ais at about 0.7. The graph of f 00 [sin( +sin)] ( + cos ) +cos( +sin)( 9sin) is even more interesting near this -value: it seems to just touch the -ais.

39 SECTION 4.6 GRAPHING WITH CALCULUS AND CALCULATORS 49 If we zoom in on this place on the graph of f 00,weseethatf 00 actually does cross the ais twice near 0.65, indicating a change in concavity for a very short interval. If we look at the graph of f 0 onthesameinterval,weseethatit changes sign three times near 0.65, indicating that what we had thought was a broad etremum at about 0.7actually consists of three etrema (two maima and a minimum). These maimum values are roughly f(0.59) and f(0.68), and the minimum value is roughly f(0.64) There are also a maimum value of about f(.96) and minimum values of about f(.46) 0.49 and f(.7) 0.5. The points of inflection on (0,π) are about (0.6, ), (0.66, ), (.7, 0.7), (.75, 0.77), and(.8, 0.4). On(π, π), they are about (4.0, 0.4), (4.54, 0.77), (5., 0.7), (5.6, ),and(5.67, ). TherearealsoIPat(0, 0) and (π, 0). Note that the function is odd and periodic with period π, and it is also rotationally symmetric about all points of the form ((n +)π, 0), n an integer.. f() 4 + c + c. Note that f is an even function. For c 0, the only -intercept is the point (0, 0). We calculate f 0 () 4 +c 4 + c f 00 () +c. Ifc 0, 0is the only critical point and there is no inflection point. As we can see from the eamples, there is no change in the basic shape of the graph for c 0;itmerely becomes steeper as c increases. For c 0, the graph is the simple curve y 4. For c<0,thereare-intercepts at 0 and at ± c. Also, there is a maimum at (0, 0),andthereare minima at ± c, 4 c.asc,the-coordinates of these minima get larger in absolute value, and the minimum points move downward. There are inflection points at ± 6 c, 56 c, whichalsomoveawayfromtheoriginasc.. Note that c 0is a transitional value at which the graph consists of the -ais. Also, we can see that if we substitute c for c, the function f() c will be reflected in the -ais, so we investigate only positive values of c (ecept c,asa +c demonstration of this reflective property). Also, f is an odd function. lim f() 0,soy 0is a horizontal asymptote ±

40 50 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION for all c. We calculate f 0 () ( + c )c c(c ) ( + c ) c(c ) ( + c ). f 0 () 0 c 0 ±/c. So there is an absolute maimum value of f(/c) and an absolute minimum value of f( /c).these etrema have the same value regardless of c, but the maimum points move closer to the y-ais as c increases. f 00 () ( c )( + c ) ( c + c)[( + c )(c )] ( + c ) 4 ( c )( + c )+(c c)(4c ) c (c ) ( + c ) ( + c ) f 00 () 0 0or ± /c,sothereareinflection points at (0, 0) and at ± /c, ± /4.Again,they-coordinate of the inflection points does not depend on c,butasc increases, both inflection points approach the y-ais. 5. f() c +sin f 0 () c +cos f 00 () sin f( ) f(),sof is an odd function and its graph is symmetric with respect to the origin. f() 0 sin c,so0is always an -intercept. f 0 () 0 cos c, so there is no critical number when c >. If c, then there are infinitely many critical numbers. If is the unique solution of cos c in the interval [0,π], then the critical numbers are nπ ±, where n ranges over the integers. (Special cases: When c, 0;whenc 0, π ;andwhenc, π.) f 00 () < 0 sin >0, sof is CD on intervals of the form (nπ, (n +)π). f is CU on intervals of the form ((n )π, nπ). Theinflection points of f are the points (nπ, nπc),wheren is an integer. If c, thenf 0 () 0 for all,sof is increasing and has no etremum. If c,thenf 0 () 0 for all,sof is decreasing and has no etremum. If c <, thenf 0 () > 0 cos > c is in an interval of the form (nπ, nπ + ) for some integer n. These are the intervals on which f is increasing. Similarly, we find that f is decreasing on the intervals of the form (nπ +, (n +)π ).Thus,f has local maima at the points nπ +,wheref has the values c(nπ + )+sin c(nπ + )+ c,andf has local minima at the points nπ,wherewehavef(nπ )c(nπ ) sin c(nπ ) c. The transitional values of c are and. Theinflection points move vertically, but not horizontally, when c changes. When c, there is no etremum. For c <, the maima are spaced π apart horizontally, as are the minima. The horizontal spacing between maima and adjacent minima is regular (and equals π)whenc 0,but the horizontal space between a local maimum and the nearest local minimum shrinks as c approaches.

41 SECTION 4.7 OPTIMIZATION PROBLEMS 5 7. (a) f() c 4 +.Forc 0, f() +, a parabola whose verte, (0, ), is the absolute maimum. For c>0, f() c 4 +opens upward with two minimum points. As c 0, the minimum points spread apart and move downward; they are below the -ais for 0 <c< and above for c>. Forc<0, the graph opens downward, and has an absolute maimum at 0and no local minimum. (b) f 0 () 4c 4 4c( /c) [c 6 0]. If c 0, 0 is the only critical number. f 00 () c 4,sof 00 (0) 4 and there is a local maimum at (0,f(0)) (0, ),whichliesony.ifc>0, the critical numbers are 0 and ±/ c. As before, there is a local maimum at (0,f(0)) (0, ),whichliesony. f ±/ 00 c 48> 0, so there is a local minimum at ±/ c.heref ±/ c c(/c ) /c + /c +. But ±/ c, /c + lies on y since ±/ c /c. 4.7 Optimization Problems. (a) First Number Second Number Product We needn t consider pairs where the first number is larger than the second, since we can just interchange the numbers in such cases. The answer appears to be and, but we have considered only integers in the table. (b) Call the two numbers and y. Then + y,soy. Call the product P.Then P y ( ),sowewishtomaimizethefunctionp().sincep 0 (), we see that P 0 () 0.5. Thus, the maimum value of P is P (.5) (.5).5 and it occurs when y.5. Or: Note that P 00 () < 0 for all,sop is everywhere concave downward and the local maimum at.5 must be an absolute maimum.. The two numbers are and 00 00,where>0. Minimize f() +. f 0 () The critical number is 0.Sincef 0 () < 0 for 0 <<0 and f 0 () > 0 for >0, there is an absolute minimum at 0. The numbers are 0 and 0.

42 5 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION 5. If the rectangle has dimensions and y, thenitsperimeteris +y 00 m, so y 50. Thus, the area is A y (50 ). We wish to maimize the function A() (50 ) 50,where0 <<50. Since A 0 () 50 ( 5), A 0 () > 0 for 0 <<5 and A 0 () < 0 for 5 <<50. Thus,A has an absolute maimum at 5,andA(5) 5 65m. The dimensions of the rectangle that maimize its area are y 5m. (The rectangle is a square.) 7. We need to maimize Y for N 0. Y (N) kn +N Y 0 (N) ( + N )k kn(n) k( N ) k( + N)( N). Y 0 (N) > 0 for 0 <N<and Y 0 (N) < 0 ( + N ) ( + N ) ( + N ) for N>. Thus, Y has an absolute maimum of Y () k at N. 9. (a) Theareasofthethreefigures are,500,,500,and9000 ft. There appears to be a maimum area of at least,500 ft. (b) Let denote the length of each of two sides and three dividers. Let y denote the length of the other two sides. (c) Area A length width y (d) Length of fencing y 750 (e) 5 +y 750 y 75 5 A() (f) A 0 () SinceA 00 () 5 < 0 there is an absolute maimum when 75.Then y Thelargestareais ,06.5 ft. These values of and y are between the values in the first and second figures in part (a). Our original estimate was low.. y.5 0 6,soy /. Minimize the amount of fencing, which is +y +( /) / F (). F 0 () 0 6 / ( 0 6 )/. The critical number is 0 and F 0 () < 0 for 0 <<0 and F 0 () > 0 if >0, so the absolute minimum occurs when 0 and y.5 0. The field should be 000 feet by 500 feet with the middle fence parallel to the short side of the field.. Let b be the length of the base of the bo and h theheight.thesurfaceareais00 b +4hb h (00 b )/(4b). The volume is V b h b (00 b )/4b 00b b /4 V 0 (b) 00 4 b. V 0 (b) b b 400 b SinceV 0 (b) > 0 for 0 <b<0 and V 0 (b) < 0 for b>0, there is an absolute maimum when b 0by the First Derivative Test for Absolute Etreme Values (see page 59). If b 0,thenh (00 0 )/(4 0) 0, so the largest possible volume is b h (0) (0) 4000 cm.

43 SECTION 4.7 OPTIMIZATION PROBLEMS (w)(w)h w h,soh 5/w.Thecostis C(w)0(w ) + 6[(wh)+hw]+6(w ) w +6wh w + 80/w C 0 (w) 64w 80/w 4(6w 45)/w w 45 is the critical number. 6 C0 (w) < 0 for 0 <w< 45 and 6 C 0 (w) > 0 for w> 45. The minimum cost is C 45 (.85) / $ The distance from a point (, y) on the line y 4 +7to the origin is ( 0) +(y 0) + y.however,itis easier to work with the square of the distance; that is, D() + y + y +(4 +7). Because the distance is positive, its minimum value will occur at the same point as the minimum value of D. D 0 () +(4 + 7)(4) 4 +56,soD 0 () D 00 () 4> 0,soD is concave upward for all. Thus,D has an absolute minimum at 8 7. The point closest to the origin is (, y) 8, , From the figure, we see that there are two points that are farthest away from A(, 0). The distance d from A to an arbitrary point P (, y) on the ellipse is S d ( ) +(y 0) and the square of the distance is S d ++y ++(4 4 ) +5. S 0 6 and S 0 0.NowS00 6 < 0, so we know that S has a maimum at.since, S( ) 4, 6,andS() 0, we see that the maimum distance is 6. The corresponding y-values are y ± 4 4 ± ± 4 9 ±.89. The points are, ± 4.. The area of the rectangle is ()(y) 4y. Alsor + y so y r and y r. r r y r, so the area is A() 4 r.now A 0 () 4 r r. Clearly this gives a maimum. 4 r. The critical number is r r r, which tells us that the rectangle is a square. The dimensions are r

44 54 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION. The height h of the equilateral triangle with sides of length L is L, since h +(L/) L h L 4 L 4 L h L. Using similar triangles, L y L L/ L y y L y (L ). The area of the inscribed rectangle is A() ()y (L ) L,where0 L/. Now 0A 0 () L 4 L 4 L/4. SinceA(0) A(L/) 0, the maimum occurs when L/4,andy L 4 L 4 L, so the dimensions are L/ and 4 L. 5. Theareaofthetriangleis A() (t)(r + ) t(r + ) r (r + ). Then 0A 0 () r r + r + r + r r + r + r r r + r r 0 + r r ( r)( + r) r or r. NowA(r) 0A( r) the maimum occurs where r, so the triangle has height r + r r and base r r 4 r r. 7. The cylinder has volume V πy (). Also + y r y r,so V () π(r )() π(r ),where0 r. V 0 () π r 0 r/.nowv (0) V (r) 0,sothereisa maimum when r/ and V r/ π(r r /) r/ 4πr /. 9. Thecylinderhassurfacearea (area of the base) + (lateral surface area) π(radius) +π(radius)(height) πy +πy() Now + y r y r y r, so the surface area is S()π(r )+4π r, 0 r πr π +4π r

45 Thus, S 0 () 0 4π +4π 4π (r ) / ( )+(r ) / SECTION 4.7 OPTIMIZATION PROBLEMS 55 r + r 4π r + r r S 0 () 0 r r () r (r ) (r )r 4 4r +4 4 r 4 r 4 4r r + r 4 0. This is a quadratic equation in. By the quadratic formula, 5 ± 5 0 r, but we reject the root with the + sign since it 5 doesn t satisfy (). [The right side is negative and the left side is positive.] So 5 r. SinceS(0) S(r) 0,the 0 maimum surface area occurs at the critical number and r y r r r the surface area is 5+ π 5 r +4π r πr (5 5)(5+ 5) πr πr πr πr y 84 y 84/. Total area is A() (8+)( + 84/) ( /),so A 0 () ( 56/ )0 6. There is an absolute minimum when 6since A 0 () < 0 for 0 <<6 and A 0 () > 0 for >6. When 6, y 84/6 4, so the dimensions are 4 cm and 6 cm.. Let be the length of the wire used for the square. The total area is 0 0 A() (0 ), 0 0 A 0 () (0 ) NowA(0) 6 A(0) and A 40.7,so (a ) The maimum area occurs when 0m, and all the wire is used for the square. (b) The minimum area occurs when m. 5. The volume is V πr h and the surface area is V S(r) πr +πrh πr +πr πr + V πr r. S 0 (r) πr V V r 0 πr V r π cm. This gives an absolute minimum since S 0 (r) < 0 for 0 <r< V π and S0 (r) > 0 for r> V π. When r V π, h V πr V V π(v/π) / π cm ,

46 56 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION 7. h + r R V π r h π (R h )h π (R h h ). V 0 (h) π (R h )0when h R. This gives an absolute maimum, since V 0 (h) > 0 for 0 <h< R and V 0 (h) < 0 for h> R. The maimum volume is V R π R R 9 πr. 9. By similar triangles, H R H h r so we ll solve () for h. h H Hr R HR Hr R Hr R H h (). The volume of the inner cone is V πr h, H (R r) (). R Thus, V(r) π r H πh (R r) R R (Rr r ) V 0 (r) πh R (Rr r ) πh r(r r). R V 0 (r) 0 r 0or R r r R and from (), h H R R R H R R H. V 0 (r) changes from positive to negative at r R, so the inner cone has a maimum volume of V πr h π R H 4 7 πr H, which is approimately 5% of the volume of the larger cone. 4. P (R) E R (R + r) P 0 (R) (R + r) E E R (R + r) [(R + r) ] (R +Rr + r )E E R E Rr (R + r) 4 E r E R E (r R ) E (r + R)(r R) E (r R) (R + r) 4 (R + r) 4 (R + r) 4 (R + r) P 0 (R) 0 R r P (r) E r (r + r) E r 4r E 4r. The epression for P 0 (R) shows that P 0 (R) > 0 for R<rand P 0 (R) < 0 for R>r. Thus, the maimum value of the power is E /(4r), and this occurs when R r. 4. S 6sh s cot θ +s csc θ (a) ds dθ s csc θ s csc θ cot θ or s csc θ csc θ cotθ. (b) ds dθ 0when csc θ cotθ 0 sin θ cos θ 0 sin θ cos θ. The First Derivative Test shows that the minimum surface area occurs when θ cos 55.

47 SECTION 4.7 OPTIMIZATION PROBLEMS 57 (c) If cos θ,thencot θ and csc θ, so the surface area is S 6sh s +s 6sh s + 9 s 6sh + 6 s 6s h + s Here T () 6 + 5, 0 5 T 0 () ( + 5) 5 7.But 5 7 > 5, sot has no critical number. Since T (0).46 and T (5).8,he should row directly to B. 47. There are (6 ) km over land and +4km under the river. We need to minimize the cost C (measured in $00,000) of the pipeline. C() (6 )(4) + +4 (8) C 0 () 4+8 ( +4) / 8 () C 0 () / [0 6]. Compare the costs for 0, /,and6. C(0) , C / 4 8/ +/ 4+4/ 7.9,andC(6) So the minimum cost is about $.79 million when P is 6 / 4.85 km east of the refinery. 49. The total illumination is I() k + k, 0 <<0. Then (0 ) I 0 () 6k k + (0 ) 0 6k(0 ) k (0 ) (0 ) ft. This gives a minimum since I00 () > 0 for 0 <<0. 5. Every line segment in the first quadrant passing through (a, b) with endpoints on the - and y-aes satisfies an equation of the form y b m( a),wherem<0. By setting 0and then y 0,wefind its endpoints, A(0,b am) and B a b, 0.The m distance d from A to B is given by d [ a b m 0] +[0 (b am)]. It follows that the square of the length of the line segment, as a function of m,isgivenby S(m) a b +(am b) a ab m m + b m + a m abm + b. Thus,

48 58 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION S 0 (m) ab m b m +a m ab m (abm b + a m 4 abm ) m [b(am b)+am (am b)] m (am b)(b + am ) Thus, S 0 (m) 0 m b/a or m b.sinceb/a > 0 and m<0, m must equal b.since a a m < 0,wesee that S 0 (m) < 0 for m< b and a S0 (m) > 0 for m> b.thus,s has its absolute minimum value when m a That value is S b a a + b a + a b b a + ab b a + a b + b a +a 4/ b / + a / b 4/ + a 4/ b / +a / b 4/ + b a +a 4/ b / +a / b 4/ + b The last epression is of the form + y +y + y [( + y) ] with a / and y b /, so we can write it as (a / + b / ) and the shortest such line segment has length S (a / + b / ) /. 5. (a) If c() C(), then, by Quotient Rule, we have c0 () C0 () C().Nowc 0 () 0when C 0 () C() 0 and this gives C 0 () C() c(). Therefore, the marginal cost equals the average cost. (b) (i) C() 6, /, C(000) 6, , , , ,49, so C(000) $4,49. c() C()/ 6, /, c(000) $4.49/unit. C 0 () /, C 0 (000) $89.74/unit. (ii) We must have C 0 () c() / 6, / / 6,000 (8,000) / 400 units. To check that this is a minimum, we calculate c 0 () 6,000 + (/ 8000).Thisisnegativefor<(8000) / 400,zeroat 400, and positive for > 400,so c is decreasing on (0, 400) and increasing on(400, ). Thus,c has an absolute minimum at 400. [Note: c 00 () is not positive for all >0.] (iii) The minimum average cost is c(400) $0/unit. 55. (a) We are given that the demand function p is linear and p(7,000) 0, p(,000) 8, so the slope is 0 8 and an equation of the line is y 0 7,000, ( 7,000) y p() +99 (/000). 000 (b) The revenue is R() p() 9 ( /000) R 0 () 9 (/500) 0 when 8,500. Since R 00 () /500 < 0, the maimum revenue occurs when 8,500 the price is p(8,500) $9.50. b. a

49 SECTION 4.7 OPTIMIZATION PROBLEMS (a) As in Eample 6, we see that the demand function p is linear. We are given that p(000) 450 and deduce that p(00) 440,sincea$0 reduction in price increases sales by 00 per week. The slope for p is so an equation is p 450 ( 000) or p() (b) R() p() R 0 () when 5(550) p(750) 75, so the rebate should be $ , (c) C() 68, P () R() C() , ,000, P 0 () when 000. p(000) 50. Therefore, the rebate to maimize profits should be $ Here s h + b /4,soh s b /4. TheareaisA b s b /4. Let the perimeter be p,sos + b p or s (p b)/ A(b) b (p b) /4 b /4b p pb/4. Now A 0 p pb bp/4 (b) 4 p pb pb + p 4 p pb. Therefore, A 0 (b) 0 pb + p 0 b p/. SinceA 0 (b) > 0 for b<p/ and A 0 (b) < 0 for b>p/,there is an absolute maimum when b p/.but then s + p/ p,so s p/ s b the triangle is equilateral. 6. Note that AD AP + PD 5 + PD PD 5. Using the Pythagorean Theorem for PDB and PDC gives us L() AP + BP + CP + (5 ) + + (5 ) L 0 () From the graphs of L 0 +4 and L 0, it seems that the minimum value of L is about L(.59) 9.5 m. 6. Thetotaltimeis T ()(time from A to C)+(time from C to B) a + b +(d ) +, 0 <<d v T 0 () v a + v d sin θ v b +(d ) v sin θ v The minimum occurs when T 0 () 0 [Note: T 00 () > 0] sin θ v sin θ v.

50 60 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION 65. y + z, but triangles CDE and BCA are similar, so z/8 / 4 4 z / 4. Thus, we minimize f() y +4 /( 4) /( 4), 4 < 8. f 0 () ( 4)( ) [( 4) ] ( 6) 0 ( 4) ( 4) ( 4) when 6. f 0 () < 0 when <6, f 0 () > 0 when >6, so the minimum occurs when 6in. 67. It suffices to maimize tan θ. Now Let f(t) tanθ t tan(ψ + θ) tan ψ +tanθ tan ψ tan θ t +tanθ t tan θ.so t( t tan θ) t +tanθ t (+t )tanθ tan θ t +t. t f 0 (t) +t t(6t) t +t ( + t ) ( + t ) 0 t 0 t since t 0. Nowf 0 (t) > 0 for 0 t< and f 0 (t) < 0 for t>,sof has an absolute maimum when t and tan θ / + / θ π. Substituting for t and θ in t tan(ψ + θ) gives us 6 tan ψ + π ψ π In the small triangle with sides a and c and hypotenuse W, sin θ a W and cos θ c W. In the triangle with sides b and d and hypotenuse L, sin θ d L and cos θ b.thus,a W sin θ, c W cos θ, d L sin θ,andb L cos θ,sothe L area of the circumscribed rectangle is A(θ)(a + b)(c + d) (W sin θ + L cos θ)(w cos θ + L sin θ) W sin θ cos θ + WLsin θ + LW cos θ + L sin θ cos θ LW sin θ + LW cos θ +(L + W )sinθ cos θ LW (sin θ +cos θ)+(l + W ) sinθ cos θ LW + (L + W )sinθ, 0 θ π This epression shows, without calculus, that the maimum value of A(θ) occurs when sin θ θ π θ π. So the maimum area is A π 4 4 LW + (L + W ) (L +LW + W ) (L + W ).

51 SECTION 4.8 NEWTON S METHOD 6 7. (a) If k energy/km over land, then energy/km over water.4k. So the total energy is E.4k k( ), 0, and so de d.4k k. (5 + / ) Set de d 0:.4k k(5 + ) / Testing against the value of E at the endpoints: E(0).4k(5) + k 0k, E(5.) 7.9k, E() 9.5k. Thus, to minimize energy, the bird should fly to a point about 5. km from B. (b) If W/L is large, the bird would flytoapointc that is closer to B than to D to minimize the energy used flying over water. If W/L is small, the bird would flytoapointc that is closer to D than to B to minimize the distance of the flight. E W L( ) de d W L 0when W L. By the same sort of argument as in part (a), this ratio will give the minimal ependiture of energy if the bird heads for the point km from B. 5 + (c) For flight direct to D,, so from part (b), W/L.07. There is no value of W/L for which the bird should fly directly to B. But note that lim, so if the point at which E is a minimum is close to B,then 0 +(W/L) W/L is large. (d) Assuming that the birds instinctively choose the path that minimizes the energy ependiture, we can use the equation for de/d 0from part (a) with.4k c, 4,andk : c(4) (5 + 4 ) / c 4/ Newton's Method. (a) The tangent line at intersects the -ais at.,so.. The tangent line at.intersects the -ais at,so.0. (b) 5would not be a better first approimation than since the tangent line is nearly horizontal. In fact, the second approimation for 5appears to be to the left of.. Since and y 5 4is tangent to y f() at, we simply need to find where the tangent line intersects the -ais. y f() + 4 f 0 () +,so n+ n n + n 4.Now n (.) +(.) (.) +

52 6 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION 7. f() 5 f 0 () 5 4, so n+ n 5 n n.now 5 4 n (.5)5.5 5(.5) 4 9. f() + + f 0 () +,so n+ n n + n +. n + Now ( ) +( ) + ( ) Newton s method follows the tangent line at (, ) down to its intersection with the -ais at (.5, 0), giving the second approimation To approimate 5 0 (so that 5 0), wecantakef() 5 0. Sof 0 () 5 4, and thus, n+ n 5 n 0.Since 5 and is reasonably close to 0, we ll use 5 4. We need to find approimations n until they agree to eight decimal places..85,.84864, , So , to eight decimal places. Here is a quick and easy method for finding the iterations for Newton s method on a programmable calculator. (The screens shown are from the TI-84 Plus, but the method is similar on other calculators.) Assign f() 5 0 to Y,andf 0 () 5 4 to Y.Nowstore in X andthenenterx Y /Y X to get.85. By successively pressing the ENTER key, you get the approimations, 4,... In Derive, load the utility file SOLVE. EnterNEWTON(ˆ5-0,,) and then APPROXIMATE to get [,.85,.84864, , ]. You can request a specific iteration by adding a fourth argument. For eample, NEWTON(ˆ5-0,,,) gives [,.85,.84864]. In Maple, make the assignments f : ˆ5 0;, g : f()/d(f)();,and :.;. Repeatedly eecute the command : g(); to generate successive approimations. In Mathematica, make the assignments f[_ ]:ˆ5 0, g[_ ]: f[]/f 0 [],and. Repeatedly eecute the command g[] to generate successive approimations.. f() f 0 () n+ n 4 n n +5 n 6. We need to find 4 n 6 n +0 n approimations until they agree to si decimal places. We ll let equal the midpoint of the given interval, [, ]..5.65,.8808, 4.756, So the root is.756 to si decimal places.