National University of Singapore Department of Mathematics

Transcription

1 National University of Singapore Department of Mathematics Semester I, 2002/2003 MA505 Math I Suggested Solutions to T. 2. Let f() be a real function defined as follows: sin(a) if <0, f() = b + c if 0, 2 if >, where a, b, c and d are real numbers. (a) Find all conditions on a, b, c such that f is continuous on (, ). (b) Find conditions on b, c and d such that f is differentiable at =. SOLUTION. (a) Obviously f is continuous at 0,. f is continuous at =0 f() =f(0) + sin(a) = c +(b + c) sin(a) a = c a a = c. Here we use the it sin t t 0 t =. f is continuous at = f() =f() f() + (b + c) =b + c b + c = d. Therefore, f is continuous in (, ) if and only if a = c, b + c = d. + 2 (b) f is differentiable at = if and only if the following two conditions are satisfied: (i) f is continuous at =, and (ii) two one-sided its f() f() f() f() and eist and equal. We have + f() f() b + c (b + c) = b.

2 On the other hand, f() f() + 2 (b + c) + 2 d by above + d( )( +) =2d. + Recall that if f is differentiable at =,thenf is continuous at =. Hence f is differentiable at =ifandonlyifd = b + c and b =2d. It follows that c = d as well. 2. For the following functions, find y. (a) 2/3 + y 2/3 = a 2/3. Differentiating the equality we get 2 3 /3 + 2 dy y /3 3 =0. Since 0 <<awe have dy y/3 a = = 2/3 2/3 = ( a /3 /3 )2/3 ; d 2 y = 2 2 ( a ( 2 3 )a2/3 5/3 a 2/3 = )2/3 3 ( 5/3 a )2/3 = a 2/3 3 4/3 a 2/3 2/3. (b) y =(sin),0<< π,sosin>0. 2 y ln y = sin ln, =cosln +cos, y y = y( + ln )cos, [ ] y = y ( + ln )cos + y ( + ln )( )+ cos2 [ cos = y( + ln ) 2 cos 2 2 ] + y ( + ln )sin. Hence y = (sin) ( + ln )cos, [ ] y = (sin) ( + ln ) 2 cos 2 + cos2 ( + ln )sin. (c) = a cos t, y = a sin t. dy = dy dt dt = a cos t a sin t d 2 y 2 = d (dy )= d ( dy ) dt dt = cot t, = d ( cot t) dt dt = sin 2 t a sin t = a sin 3 t.

3 General Remarks. We apply the chain rule quite a few times in this tutorial. Let s recall the chain rule df g() =[f g()] =[f(g())] = f (g()) g (). 3. Let y = f(u) andu = g(). Using the chain rule, show that d 2 y 2 = dy du d2 u 2 + d2 y du 2 ( ) 2 du. We assume that f and g are twice differentiable functions so that the above formula makes sense. SOLUTION: By the chain rule, we have d ( dy dy = dy du du ) = d ( dy du du ) = d ( ) dy du = d du = d2 y du 2 du + dy du d ( ) du ( ) dy du du du + dy du d2 u 2 ( ) 2 du + dy du d2 u. 2 (product rule) 4. Assume f has continuous derivatives up to the second order. Find y. (a) y = f( 2 ), (b) y = f ( ), y = 2f ( 2 ), y = 2f ( 2 )+4 2 f ( 2 ).. y = 2 f ( ), y = 2 3 f ( )+ 4 f ( ). (c) y = f(ln ), y = f (ln ), y = 2 f (ln )+ 2 f (ln ).

4 (d) y = f(e ), y = e f (e ), y = e f (e )+e 2 f (e ). 5. Show that the tangents to the curve y = π at = π and = π intersect at right angle. SOLUTION. y = π, y cos = π. 2 At = π we have y =0,y =. The tangent line l at = π is given by y = π. At = π we have y =0,y =. The tangent line l 2 at = π is given by y = π +. l and l 2 intersect at (0,π). The slopes of l, l 2 are k = andk 2 = respectively. k k 2 =. Hence l and l 2 intersect at right angle. Or else l : + y = π = the vector, is to the line l ; l 2 : ( ) + y = π = the vector, is to the line l 2. By finding the dot product,, = + = 0, we can also see out that l l Consider differentiable functions f and g on IR. Suppose that f is increasing on IR, and g is decreasing on IR. Are the following statements always true? (a) f g is increasing on IR. (b) f g is decreasing on IR. Justify your answers. SOLUTION. (a) f is differentiable and increasing f () > 0;g is differentiable and decreasing g () < 0. By the product formula, (f g) () =f ()g()+g ()f(). But we CANNOT conclude that (f g) () > 0 for all IR (i.e., f g is increasing on IR). For eample, f() =, which is differentiable and increasing; and g() =, which is differentiable and decreasing; but f g = 2 is NOT increasing on IR. It follows that statement (a) is NOT always true. (b) By the chain rule, (f g) () =f (g()) g () < 0 as f (y) > 0, g () < 0 for all, y IR. Hence f g is decreasing on IR. Hence statement (b) is always valid.

5 7. Find the etremal values of the following functions. (7a) y = + 2 +, [ 3, 3]. y = y 2 ( +)2 ( 2 +) 2. y =0if = ± 2. < 0 if < 2, =0 if = 2, > 0 if 2 << + 2, =0 if = + 2, < 0 if > + 2. Hence y is decreasing in [ 3, 2), increasing in ( 2, + 2), and decreasing in ( + 2, 3]. y( 2) = 2( 2+), y( + 2) = 2( 2+) < 5 < 2 5 < 2( 2 ). 2( 2 ), y( 3) = 5, y(3) = 2 5. So y has an absolute minimum value at = 2 with min y = [ 3,3] 2( 2+) and has an absolute maimum value at = + 2with ma y = [ 3,3] 2( 2 ).

6 (7b) y =( ) 3 2, (, ). When 0wehave y = 2/ ( ) /3 = /3. y does not eist at = 0. The critical points are =0an = 2 5. y > 0 if <0, does not eist if =0, < 0 if 0 << 2, 5 =0 if = 2, 5 > 0 if > 2. 5 Hence y is increasing in (, 0), decreasing in (0, 2), and increasing in ( 2, ). 5 5 = 0 is a local maimum point with y(0) = 0. = 2 is a local minimum point with y( 2)= 3( )2/3. y =, y =. Soy has no absolute etremes.

7 (7c) y = ln(+), (, ). For > wehavey = + = +, y < 0in(, 0) and y > 0in(0, ). The only critical point is =0. y is decreasing in (, 0) and increasing in (0, ). y =, y =. + Therefore y has an absolute minimum value at =0with has no absolute maimum value. min y = y(0) = 0. y <<

8 (7d) y =2tan tan 2, [0, π 2 ). y = 2 cos 2 2tan cos 2 = 2 ( tan ). cos 2 y > 0if0 < π, 4 y < 0if π << π. The only critical point is = π. y is increasing in [0, π) and decreasing in ( π, π ). y(0) = 0, y( π ) = and π 2 y =. Therefore y has a local minimum value at =0. y has an absolute maimum value at = π with ma y =.yhas no absolute minimum value. 4 0 < π 2

9 (7e) y =sin 3 +cos 3, y =3sin 2 cos 3cos 2 = 3 sin 2( cos ). 2 y =0if = kπ, kπ + π,orkπ + π for any integer k Z. Here Z = { 4 2 2,, 0,, 2 } is the set of all integers. y =3cos2( cos )+ 3 sin 2( +cos). 2 { > 0 if k is odd,. y (kπ) = 3coskπ =3( ) k+ So y has a local maimum value at =2nπ with y(2nπ) =, and a local minimum value at =(2n+)π < 0 if k is even. with y((2n +)π) = for any n Z. 2. y (kπ + π 4 )=3sin(kπ + π { 4 )=3 2 > 0 if k is even, 2 ( )k So y has a local < 0 if k is odd. maimum value at =(2n +)π + π with y((2n +)π + π )= 2,andalocal minimum value at =2nπ + π with y(2nπ + π )= 2 for any n Z y (kπ + π 2 )=3cosπsin(kπ + π { > 0 if k is odd, 2 )=3( )k+ So y has a < 0 if k is even. local maimum value at =2nπ + π with y(2nπ + π ) =, and a local minimum 2 2 value at =(2n +)π + π with y((2n +)π + π )= for any n Z. 2 2 Summarizing, we have: y has absolute maimum values at =2nπ, and2nπ + π,with ma y =, and 2 << a local maimum value at =(2n +)π + π with y((2n +)π + π )= 2, for any n Z. y has absolute minimum values at =(2n+)π and (2n+)π + π with min y = 2 <<, and a local minimum value at =2nπ+ π with y(2nπ+ π)= 2, for any n Z

10 8. Given a constant c, eplain why the equation c =0hasatleastone real root. For what values of c does the equation have only one real root? Solution. Set f() = c. Forafiednumberc, clearly when > 0islarge, then f( ) is positive; and when 2 < 0isvery negative, thenf( 2 ) is negative. The function is continuous on IR. It follows from the Intermediate Value Theorem that there is a real number o such that f( o )= 3 o 3 o + c =0. That is, the equation has at least one real root. f () =3( 2 ) > 0 if <, =0 if =, < 0 if <<, =0 if =, > 0 if >. Thus f is increasing in (, ), decreasing in (, ) and increasing in (, ). It follows from the above (or the second derivative test) that = iswheref achieves a local maimum, an =iswheref achieves a local minimum. Furthermore, ( ) f( ) = c +2 (localma), f() = c 2 (local min). An useful insight is that adding the number c to the epression 3 3 amounts to uplifting or pushing down the graph of y = 3 3, according to c>0orc<0 (compare with the figures below). Hence if c is so large that the local minimum in (*) is also positive, then we have just one real solution. And the same if c is so negative that the local maimum is made negative. So if c>2, then f() > 0andf has only one real solution. When c =2,f() = 0 and f( ) > 0, hence f has more than one real root. When 2 <c 2and f( ) > 0 >f(), and f has more than one real root. When c = 2, we have f( ) = 0 and f has more than one real root. Finally, when c< 2,f( ) < 0 and f() < 0, and again we have only one root. See the figures for the cases c =3, 2, 0, 2 & 3. c = 3

11 c=2 c=0

12 c=-2 c=-3

13 9. Use L Hopital s rule to find the following its. (9a) π/2 +cos2 π/2 cos 2sin2 π/2 4cos2 = 4. (9b) ln(cos a) ln(cos b) a sin a cos a b sin b cos b a sin a cos b b sin b cos a = a2 b 2. (9c). (9d) (9e) (9f) tan + a ln + ln Using (4d) we have tan( ) ln a + ln 2 sec 2 ( ) 2 a a + cos 2 ( )= a a =0. =. So = e. ln + + ln + So + = e 0 =. ln + + ln =0. (9g) ( + ) =. (9h) ( ) ln 2 = 2 = 6 = 6. So ( ) 2 = e /6. ln( ) 2 cos 3 ( = 2 cos ) 2 2 cos cos 3 2