Pre-Calc Homework Solutions 118

# Pre-Calc Homework Solutions 118 - 118 Chapter 3 Review 2x...

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Unformatted text preview: 118 Chapter 3 Review 2x ln 2, f (0) f(h) h f(0) 49. (a) Since f (x) (b) f (0) lim h0 20 ln 2 2h h 20 ln 2. lim h0 s Chapter 3 Review Exercises 2h h (pp. 172175) 1 lim h0 1. 2. 3. dy dx dy dx dy dx d 5 x dx d (3 dx 1 2 x 8 1 x 4 5x 4 21x 2 1 x 4 1 4 (c) Since quantities in parts (a) and (b) are equal, lim h0 2h h 7x 3 3x 7) 21x 6 1 ln 2. (d) By following the same procedure as above using g(x) 7x, we may see that lim h0 7h h 1 ln 7. d (2 sin x cos x) dx d 2(sin x) (cos x) dx 2(cos x) (sin x) d dx 50. Recall that a point (a, b) is on the graph of y e x if and only if the point (b, a) is on the graph of y ln x. Since there are points (x, e x) on the graph of y e x with arbitrarily large x-coordinates, there will be points (x, ln x) on the graph of y ln x with arbitrarily large y-coordinates. 51. (a) The graph y4 is a horizontal line at y a. 4. 5 5. (b) The graph of y3 is always a horizontal line. a y3 2 3 4 2 sin 2 x Alternate solution: dy dx 2 cos 2 x d (2 sin x cos x) dx d sin 2x dx (cos 2x)(2) 2 cos 2x dy dx ds dt ds dt d 2x dx 2x 1 1 (2x 1)(2) (2x (2x 1)2 1)(2) (2x 4 1)2 0.693147 1.098613 1.386295 1.609439 d cos (1 dt 2 d cot t dt 2 2 2 csc t t2 d dx 1 1/2 x 2 2t) csc 2 sin (1 2 d 2 t dt t 2t)( 2) csc 2 2 sin (1 2 t 2 t2 2t) ln a 0.693147 1.098612 1.386294 1.609438 6. We conclude that the graph of y3 is a horizontal line at y ln a. (c) d x a dx a x if and only if y3 y2 So if y3 ln a (d) y2 y1 d x ln a, then a will equal a x if and only if dx 1. 7. dy dx x 1 1 3/2 x 2 1 1, or a d x a dx e. a x ln a. This will equal y1 1, or a x and e. d (ln x dx d 1/2 (x dx 1 1 2x3/2 2 x x 1 2x 1 1 x 1/2 ) a x if and 8. dy dx only if ln a 52. d dx 1 2 x 2 d (x 2x 1) (x) dx 2 x (2x 1) 3x 1 2x 1 2x 1 (2) ( 2x 1)(1) k c) 1 . x 9. dr d d sec (1 d 3 ) sec (1 3 ) 3 ) tan (1 3 )(3) Therefore, at any given value of x, these two curves will have perpendicular tangent lines. 10. 53. (a) Since the line passes through the origin and has slope 1 , its equation is y e x . e dr d 3 sec (1 d tan 2 (3 d 3 ) tan (1 2 ) tan (3 2 2 2 tan (3 2 tan (3 4 tan (3 11. dy dx 2 d ) d ) 2 ) sec 2 (3 2 )( 2 ) 2 (b) The graph of y y ln x lies below the graph of the line e. Therefore, ln x x for all e ) sec (3 2 ) x for all positive x e positive x e. x or ln x e d 2 (x csc 5x) dx (x 2)( csc 5x cot 5x)(5) x. 5x 2 csc 5x cot 5x 12. 13. 14. dy dx dy dx dy dx d ln dx d ln (1 dx d (xe x) dx (csc 5x)(2x) (c) Multiplying by e, e ln x (d) Exponentiating both sides of ln x e x, we have e e ln x e x, or x e e x for all positive x e. (e) Let x to see that e 2x csc 5x x 1 1 x d (1 e x dx 2 1 x 1 1 ,x 2x ex ex e . Therefore, e is bigger. x 1 d x dx 1 x 0 e x) (x)(e e x) x )( 1) (e )(1) xe x e x ...
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