Pre-Calc Homework Solutions 135

Pre-Calc Homework Solutions 135 - Section 4.2 15. (a) f is...

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: Section 4.2 15. (a) f is continuous on [0, 1] and differentiable on (0, 1). (b) f (c) 2c 2 2c c f (1) f (0) 1 0 2 ( 1) 1 135 20. (a) The secant line passes through (1, f (1)) (3, f (3)) 2 3 0 1 1 2 (1, 0) and (3, 2 2 2), so its slope is 1 2 1 2 1 2 . 1 2 1 1 2 The equation is y or y x (x 1) 0.707x 0 0.707. . , or y 16. (a) f is continuous on [0, 1] and differentiable on (0, 1). (b) f (c) 2 1/3 c 3 f (1) f (0) 1 0 1 0 1 3 2 3 3 2 8 27 (b) We need to find c such that f (c) 1 2 c 1 1 2 c 1/3 2 c c 1 1 c f (c) 1 2 3 2 2 c c 17. (a) f is continuous on [ 1, 1] and differentiable on ( 1, 1). (b) f (c) f (1) 1 2 f 3 2 1 2 1 1 2 f ( 1) ( 1) 2 The tangent line has slope 3 1 , . Its equation is y 2 2 1 1 and passes through 2 1 2 x 3 2 1 2 or 1 1 c 2 y x , or y 0.707x 0.354. 0. 2 2 4 2 2 2 2 1 2/3 x , f is not differentiable at x 3 1 1 c2 c2 c 2 21. (a) Since f (x) (b) 4 2 1 c 1 4 2 0.771 [ 1, 1] by [ 1, 1] 18. (a) f is continuous on [2, 4] and differentiable on (2, 4). (b) f (c) 1 c 1 f (4) 4 ln 3 2 2 ln 3 2 ln 3 f (2) 2 ln 1 (c) f (c) 1 2/3 c 3 1 2/3 c 3 f (1) f ( 1) 1 ( 1) 1 ( 1) 2 c 1 c 1 3 3 3/2 1 2.820 (0.5, 2.5) 2.5. c 2/3 19. (a) The secant line passes through (0.5, f (0.5)) and (2, f(2)) (2, 2.5), so its equation is y c 0.192 (b) The slope of the secant line is 0, so we need to find c such that f (c) 0. 1 c c 2 2 c f (c) 0 1 1 f (1) 2 The tangent line has slope 0 and passes through (1, 2), so its equation is y 2. ...
View Full Document

Ask a homework question - tutors are online