Pre-Calc Homework Solutions 98

Pre-Calc Homework Solutions 98 - 98 Section 3.6 1 x cos 2 2...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 98 Section 3.6 1 x cos 2 2 1 x Since the range of the function f(x) cos is 2 2 dy 1 the largest possible value of is . dx 2 d x sin dx 2 dy dx x d x 2 dx 2 d f(g(x)) dx 53. cos (e) 1 1 , , 2 2 f (g(x))g (x) At x 2, the derivative is f (2)g (2) 1 1 ( 3) 3 f (x) 2 f (x) f (g(2))g (2) 1. 54. dy dx d (sin mx) dx (cos mx) (mx) d dx m cos mx m and passes (f) d dx f(x) 2 d f(x) f (x) dx The desired line has slope y (0) through (0, 0), so its equation is y 55. dy dx x d 2 tan 4 dx x sec 2 2 4 2 m cos 0 mx. At x f (2) 2, the derivative is 1 3 1 6(2 2) 1 12 2 2 sec 2 x d x 4 dx 4 2 f (2) 2 8 . 3d (g) 2 d 1 dx g 2(x) d [g(x)] 2 dx 2(5) ( 4)3 2[g(x)] dx g(x) 2g (x) [g(x)]3 y (1) sec 2 4 ( 2)2 . At x 3, the derivative is 10 64 1 5 . 32 d 2 [ f (x) g 2(x) dx The tangent line has slope 1, 2 tan y x 4 and passes through (x 1) 2, or (h) 1 2g (3) [g(3)]3 d dx (1, 2). Its equation is y 2. f 2(x) 1 2 f (x) f(x) f (x) f 2(x) 2 g 2(x) 2 f 2(x) g 2(x)] The normal line has slope equation is y 1 and passes through (1, 2). Its 2, or y 1 (x 1) x 1 2. g 2(x) [2f(x) d f(x) dx 2g(x) g(x)] d dx Graphical support: g(x)g (x) g 2(x) At x 2, the derivative is g(2)g (2) g 2(2) 10 3 f (2) f (2) [ 4.7, 4.7] by [ 3.1, 3.1] (8) 1 3 (2)( 3) 82 22 f 2(2) 10 3 56. (a) d [2 f (x)] dx 2 f (x) 1 2 3 2 . 3 5 3 17 At x d (b) [ f (x) dx 2, the derivative is 2 f (2) g(x)] f (x) g (x) 68 2 17 57. (a) d [5 f (x) dx g(x)] 5 f (x) g (x) At x g (3) 2 5. 5 f (1) (b) 1, the derivative is g (1) 5 d 1 3 8 3 d dx At x (c) d [ f (x) dx 3, the derivative is f (3) g(x)] f(x)g (x) 1. g(x) f (x) At x 3, the derivative is g(3) f (3) (3)(5) 15 ( 4)(2 ) 8 . d f(x)g 3(x) dx f(x) g 3(x) d dx dx g 3(x) f(x) f (x)[3g 2(x)] g(x) 3 f (x)g 2(x)g (x) At x g 3(x) f (x) g 3(x) f (x) g 3(0) f (0) f(3)g (3) 0, the derivative is 3 f(0)g 2(0)g (0) 1 3 3(1)(1)2 d f(x) (d) dx g(x) g(x) f (x) f(x)g (x) [g(x)]2 (1)3(5) 6. At x 2, the derivative is (2) 1 3 g(2) f (2) f(2)g (2) [g(2)]2 74 3 (8)( 3) (2)2 4 37 . 6 ...
View Full Document

This note was uploaded on 10/05/2011 for the course MAC 1147 taught by Professor German during the Fall '08 term at University of Florida.

Ask a homework question - tutors are online