Chapter 4 Derivatives [ ] = ( ) ( )= + ( ) + + = ()= + ()+ Exercise 4.1. Review of Prerequisite Skills. 1. f. 6. d. 4. b. lim. x x. = lim = c.

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1 Capter Derivatives Review of Prerequisite Skills. f. p p p 7 9 p p p Eercise.. i. ( a ) a ( b) a [ ] b a b ab b a. d. f. 9. c. + + ( ) ( + ) + ( + ) ( + ) ( + ) ( ) ( + ) + + ( ) ( ) ( + ) + 7 ( + ) b. + + f ; a f a+ f a f ( a) lim Since a, f( + ) ( + ) + ( + ) Now f + f 9 f lim + ( + 9) lim f () + () f lim f 9 f( + ) f ( + ) + ( + ) lim lim 9 d Capter : Derivatives 7

2 c. f( ) + ; a c. f( + ) f( ) f( ) + f ( ) lim f( + ) f( ) f ( ) lim + lim lim lim lim + lim ( ) ( + + ) f ( ) lim f ( ) d. f( ) f( + ) f( ) f ( ) lim. a. f( ) + f( + ) f( ) f ( ) lim ( + ) lim ( + ) + ( + ) ( + ) lim lim ( + ) ( ) + + lim lim lim( + + ) ( + ) ( ) f ( ) + f ( ) b. f( ) + f( + ) f( ). b. + f ( ) lim lim lim lim lim ( + + ) ( + ) ( + )( ) lim lim ( + + ) ( + ) ( + )( ) f ( ) + ( ) 7 Capter : Derivatives

3 7. Since f f f( ) lim + f( + ) ( + ) ( + ) + + f( ) f + f + + f ( ) lim lim + f ( ) Slopes of te tangents at,, and are f, f, and f. f( ) vt () t + lim 8 lim t + 8 vt () t+ 8 Velocities at t,, and are v 8, v, and v. () 9. f( ) +, parallel to + f + f f ( ) lim lim lim ( + ) lim lim f ( ) + 8. s(t) t + 8t; t, t, t st st vt () s () t lim + st ( + ) ( t+ ) + 8( t+ ) t t + 8t+ 8 + ( t + 8) st+ st t 8 Te slope of te tangent to to +. f ( ) Te point of tangenc will be 8, 8 8,. Te equation of te line will be or +. f( ) + ( f ()) is parallel 8 Capter : Derivatives 7

4 . + at (, ). f ( ) f( + ) f( ) lim + lim lim lim ( ) ( + ) ( ) ( + ) f is continuous. f () But f. (Vertical tangent) Eercise... At, f ( ). Slope of te tangent at (, ) is. f( ) For <, f( ), f( ) f ( ), < f ( ), And f eists for all R and f ( ). f a, f a + f ( a) lim f a f a But f( a) f a+ lim f( a+ ) and lim... k. f f( ) f ( ). f. ( + ) + f ( ) + 9+ ( ) Capter : Derivatives

5 k.. e. i. j. t t st () t t t t s ( t) t b. 7. d. 8. b. f( ) 7 + f ( ) + f ( ) At,,. + + At,. l d. + + At,. 7 Capter : Derivatives 77

6 9. a. at + b. c. d. Slope of te tangent at is +. Equation at P 7, + + P(., ) at P, ( ) at P, + Slope at is. ( ) + e. f. ( ) + 8 at, + + At, slope is 7 Now, ( ) or Slope at is +. Now, + ( ) ;.. A normal to te grap of a function at a point is a line tat is perpendicular to te tangent at te given point. at Slope of te tangent is 8, terefore, te slope of te normal is. 8 Equation is 7 ( +) P (, 7) 78 Capter : Derivatives

7 . Parallel to + + Slope of te line is. 8. : : Now,. No real solution. Te never ave te same slope For, +.., Te slope of te tangent at A(, ) is and at Te point is,. B, 8 is. Since te product of te slopes is, te tangents at A(, ) and B, 8 will be perpendicular. Capter : Derivatives 79

8 . Points are, and,.., slope is or ± non-real Tangents wit slope are at te points, 7. and, 8. + a. Equation of tangent from If a, +. Let te point of tangenc be Pa,a +. Now, and a. a Te slope of te tangent is te slope of AP. a a. a a a 8a a 8a aa ( ) a +, slope is or or a A(, ): 8 Point (, ): Slope is. Slope is. Equation of tangent is Equation of tangent is. ( ) or 9. From te point B(, 7): Slope of BP: Slope is a. Slope is. Equation is Equation is + 7 ( ) + 7 ( ) or 7. or +. a 8. a + is tangent to at.. a + a a a + a 8a a 8a a a ( a ) ( a+ ) a a Terefore, te point of tangenc is,. Tis point lies on te line, terefore, a a( ) + a + a 7. (, ) a 8 Capter : Derivatives

9 Let te coordinates of te points of tangenc be Aa, a., slope of te tangent at A is a Slope of PA: a a a a a a a or a Coordinates of te points at wic te tangents touc te curve are, and,.. + Pab (, ) is on te curve, terefore a, b a At a, slope is +. a a But a + b b a.. 8, tangent at A, Te slope of te tangent at A, is. Equation will be + ( ) ( + 9) or Coordinates are B(, ).. Terefore, slope is b b. a a n f, f n n Slope of te tangent at is Te equation of te tangent at (, ) is: n n n + Let, nz n n. n n f ( ) n, Te -intercept is as n, and, n n and te -intercept approaces as n, te slope of te tangent at (, ) increases witout bound, and te tangent approaces a vertical line aving equation. Capter : Derivatives 8

10 . a. c. f f, < +, does not eist., < f ( ), f, since, < since +, < < since +, since b. f ( ), >, < <, < <, < ( ), ( ), and f f f do not eist. Eercise.. c. ( ) + ( ) ( ) ( + ) + ( ) + 8( ) + + ( + ) f f ( ) f ( ), < or >, < <, < or >, and f ( ) do not eist.. e. + 7 ( + 7) At, () + ( 8)( ) 8. 8 Capter : Derivatives

11 f. +,. ( ) +, ( ) ( + )+ + At, +.. Tangent to + at (, ).. b. ( )( )+ + ( )+ ( )( ) 8 Slope of te tangent at (, ) is. Te equation is + ( ) ; ( + ) + ( + ) + ( + ) ( + ) + ()+ ( + + ) ( + ) ( ) ( ) + ( ) ( )( ) ( ) ( + ) ( + ) ( + ) + Point of orizontal tangenc is (, ). 7. b. + ( + ) ( ) + + [ ] [ ] Determine te point of tangenc, and ten find te negative reciprocal of te slope of te tangenc. Use tis information to find te equation of te normal. ( ) + ( + ) ( + + ) ( + + ) + ( + ) ( + + ) + ( + ) ( + + ) ( + ) ( ) ( ) + ( )( ) + ( )( ) ()( ) 9. a. f( ) g( ) g( ) g( ) f g g g g g g g g g g + + g( ) g( ) g ( ) + + g( ) g( ) g( ) b. f( ) ( + ) ( + ) ( + ) + n f ( ) + + n ( + ) ( + ) ( + )( + ) + + n f ( ) ()()() + () K () + ()()() K () + + ()()() nn+ f ( ) ( + ) () ( n) + n g g n n n n n n g g n n g g n n ( + n) ( + n) Capter : Derivatives 8

12 f a b c f a b () Horizontal tangent at f ( ) at a+ b Since Since ( 9, ) (, 8) a + b + c 9 a b 7 a+ b 9 a+ b a 9 a, 9 b c 8 (, 8) lies on te curve, a + b+ c 9. lies on te curve, a b + c 8. c 9 Te equation is c. f f f () Slope of te line is. 8 Point is at,. Find intersection of line and curve: + +. Substitute, a. or b., < or >, < <. f f + + or +. Let. R.S. + Since satisfies te equation, terefore it is a solution. Wen, ( )+. Intersection point is tangent to te curve. (, ). Terefore, te line is 8 Capter : Derivatives

13 Eercise.. g. + + ( + )( )( + )( ) ( ) At (, 9): 9 ( 7)( ) () Te slope of te tangent to te curve at 9, is 9.. c. d. At :, + ( + ) ( + ) ( )( ) ( ) ( )( ) ( 9)( ) ( )( ) ( 9) , ( + ) ( + ) ( + + )( ) ( ) Slope of te tangent is. Terefore, or 9 or. Points are and, 9 7,. + f( ) + ( + )( ) + f ( ) ( + ) 8 f ( ) + Since + is positive or zero for all R, 8 + grap of slope. > for + f( ) + ( ). Terefore, tangents to te do not ave a negative Capter : Derivatives 8

14 9. b.. Curve as orizontal tangents wen. No value of will give a orizontal slope, terefore, tere are no suc tangents. Population is growing at a rate of 7. bacteria per our at t and at. One bacteria per our at t. t. a. st (), t t, s( ) t+ Te boat is initiall m from te dock. b. + ( ) ( + ) ( + )( ) +, + t pt () + t + ( t + ) t( t) p ( t) ( t + ) ( t ) ( t + ) ( 9) p. 7. p ( + ) + ( + ) +. 8 t t vt () s () t ( + )( ) ( ) ( t+ ) 9 vt () t+. At t, v( ), te boat is moving towards te dock at a speed of m/s. Wen st (), te boat will be at te dock.., t. Te speed of te boat wen it bumps into te dock is m/s. For te tangents to te grap of to ave positive slopes, f ( )>. ( c + d) is positive for all R. ad bc > will ensure eac tangent as a positive slope. Eercise.. b. If g and f( ), e. f f ( ) f ( ) t t + 9 v( ) 9 ten a + b d, c + d c ( c + d)( a) a + b c c + d ad bc c + d f( g ) f( ). f ( + ) ( + + ) + If g + and f( ) ( + ), ( ) 9 ten f g g ( g + ) ( + ) f( ) 8 Capter : Derivatives

15 . f( ) and f( g( ) ) ( ) f g g g g, fg ( ) + 7 f( ) + 7 g, fg ( ) f( ) [+ ] f f( ) + f( ) is a quadratic function. Let f( ) a + b + c. ( ) ( ) + ( )+ Or, since g is linear and f g is quadratic, f g a b c Equating coefficients: a b a b 9a b + c c 9. f( ) f( ) + f( ), f( g( ) ) + 8+ But f( g( ) ) g( ) [ g ] + 8+ ( + ) g + a ba + ga + b b + c a + ( b a)+ 9a b + c [ ]. or g. + f, g and f( g u ) + gu ( ) 8 8 ( u ) (( ) ) ( ) ( ( )) and f g u f u ( ( )) Since f g u is quadratic, u must be linear. Let u a+ b. Now, ( a + b) ( a + b) a, a, or a ab a 8, b, or b b. + u or u f( ), g a. g( f( ) ) g ( ) b. f g f u ( u ) u + 8 ( ) Capter : Derivatives 87

16 . f( g( ) ) g f( ) ( + )+ ( + ) + ( + ) ( + ) ( + ) or. a. f( ) 7 f ( ) f f f f f f b. f g f( ) ( ) 7 7 (f g) Eercise. Note: g ( ). + 7 g f g f. i. 9 ( ) ( + ) ( ) ( + ) + ( + ) + 88 Capter : Derivatives

17 . f. ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( )[ ( ) ( )] ( ) ( ) ( ) ( + ) ( ) 8. 7 at z 7 () Slope of te tangent is. Equation of te tangent at (, ) is 9. k. s t t ds dt ( t ) ( 9t )( t) + ( t ) ( t) ( ) l. ( ) ( )( ) ( ) ( ) ( ) + ( ) [ ] ( t ) ( t) t ( t) ( t ) ( t ) ( t) ( 9t t + ) ( t ) ( t) ( 9t t ) 9. a. d. u u + u, u du u, du du du ( u )( ) ( 8 )( ) ( ) ( + ) u u +, u, du ( u + ) + u ( u + ) + du du [ + ( ) ][ () ] du. + ( + )( ) For te same slope, ( + ) +. f +,, f f ( + ) ( + ) f ( + ) ( ) or. Curves ave te same slope at and. Capter : Derivatives 89

18 . g ( ( ) ) (, ) + g ( ( ) ) ( ) Wen. fg, terefore.. ( + )( ) ( ) ( + ) + ( ) ( + ) 9 ( ) + At,. and 9 g f u u, g, g Now, f ( g ) g f () g ( ). Since f u u, f u u, and f, ( ) pqr a. +, (). + + p q r ( ) p q r p q r + ( + ) + 9. R ( ) f ( g( ) ) g ( ) At te point,, and te slope of te tangent will be ( + ) ( + ).. Equation of te tangent at, is. Solving tis equation wit te function, we ave ( + ) + + or Since and are bot double roots, te line wit equation will be a tangent at bot and. Terefore, is also a tangent at,. ( + ) + ( + ) ( ) ( + ) ( + ) + + ( + ) + ( + ) ( + )( ) +. If, prove n du u n nu. For n, u and true., wic is Assume te statement is true for n k, i.e., u k, k du ten u. For sow, k u du k n k+ ( + ). Now, u k + k u u. du u du 9 Capter : Derivatives

19 7. Terefore, if te statement is true for n k, it will be true for n k +. Since it is true for n, it will be true for n, terefore true for all n N. f( ) a+ b, f( g( ) ) f( c + d) ac ( + d)+ b ac + ad + b g( f( ) ) g( a + b) ca ( + b)+ d ac + bc + d Now, du k k du u + u ku du k k du u + k u du k u ( k+ ) k u du k ( + ) f( g( ) ) g( f( ) ). ac + ad + b cc + bc + d ad d bc b da bc ( ) If f g g f, ten da bc. Review Eercise. f. ( ) + ( ) g c+ d.. c. f ( ) + ( + ) ( ) ( ) ( + ) ( + ) ( + ) ( ) [ + + ] + ( 9 ) + ( ) i. ( ) + + ( + )( ) ( ) + ( + ) + + ( + ) ( ) ( + ) ( + ) ( + + ) ( + ) ( + + ) 8 ( ) Capter : Derivatives 9

20 . a. g f g f b. 7. b. + f f f u + +, u, u u du u u ( ) + ( u ) 8 u 8 du u du + du ± + ( ) ( + ) ±, ±. a. i). b. Horizontal tangent,, ±, ±. at, A+ ()( ) Equation of te tangent at (, ) is +. c. f + 9, f, + f f () Slope of + is. 8 9 Capter : Derivatives

21 Since perpendicular, 7 f( ) crosses te -ais at, and +. Equation of te tangent at (, ) is ( ) 7. f f. 8+ b is tangent to Slope of te tangent is 8, terefore 8,. Point of tangenc is (, 8). Terefore, 8 + b, b 8. Or 8 + b 8 b 8 8b ± +. For tangents, te roots are equal, terefore + 8b, b 8. Point of tangenc is (, 8), b b. To find a, let f( ). Terefore a. C a. f( ) a. C ( ) + f ( ) [ ] f or 9. b. C Production level is gloves/week. R 7 a. Marginal Revenue R ( ) 7 b. ( ) R 7 $. 7 Capter : Derivatives 9

22 . Dp p > p, D ( p) ( p ) ( p ) D ( ) 8. a. b. + π + ( 9) 9 9 Slope of demand curve at (, ) is. Capter Test. f is te grap on te rigt and below te -ais (it s a cubric). f is te oter grap (it is quadratic). c f + f f( ) lim + ( + ) ( ) lim + ( + + ) + lim lim ( ) lim lim d Terefore,. d. + + ( + ) ( + ) e. 7 ( 7) + ( ) ( ) 7 ( 7) (( 7)+ ) ( 7) ( ) Capter : Derivatives

23 . f ( + )( 7 )+ + At 8,, 7. At,, ()( ). Equation of tangent line at + +. (, ) is ()()+ ( ). 8. Pt () t + Te slope of te tangent to at 8, is. + 7 P P ( t) t + t +. u + u u + du u + du ( + ) ( u + ) Te amount of pollution is increasing at a rate of 7 p.p.m./ear. At, u. Capter : Derivatives 9

24 9. Normal line as a slope of. Terefore,. Terefore, as a normal line wit a slope of at,.. + a+ b + a Since te parabola and cubic function are tangent at (, ), ten + a. At (, ) a a. Since (,) is on te grap of + + b, + + b b. ()+ () Te required values are and for a and b respectivel.. + For a orizontal tangent line,. + or Te required points are,,(, ). 7 9 Capter : Derivatives