16A-MT2Solution

16A-MT2Solution - 1 Mat16A-MT2-5/14/10 1 2 3 TOT 4+4 pt 4+4...

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Unformatted text preview: 1 Mat16A-MT2-5/14/10 1 2 3 TOT 4+4 pt 4+4 pt 4 pt /20 Name: Please write your solution on this paper. 1. Describe the x-values at which the fcts are differentiable and explain why A: | x 2- 4 | x 2- 4 is positive for x 2 and x - 2. It is negative when- 2 < x < 2. Thus f ( x ) = | x 2- 4 | = ( x 2- 4 if x 2 , x - 2,- ( x 2- 4) if- 2 < x < 2. Thus f ( x ) = ( 2 x if x > 2 , x <- 2,- 2 x if- 2 < x < 2. Thus the equation is differentiable for all values of x > 2 , x <- 2 ,- 2 < x < 2. Now note that lim x - 2- f = 4 and lim x - 2 + f =- 4 thus the derivative does not exist at x =- 2 since the limits of the derivative from the left and right are not equal at x =- 2. Similarly the derivative does not exist at x = 2. Thus the derivative exists everywhere except at x = 2 ,- 2. B: ( x + 1) 2 3 For f ( x ) = ( x +1) 2 3 then f ( x ) = 2 3 ( x +1)- 1 3 = 2 3 1 ( x +1) 1 3 . Thus f exists everywhere except x =- 1 where it is undefined. Since it is undefined there, then f ( x ) cannot...
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This note was uploaded on 10/08/2010 for the course MATH 16A taught by Professor Sabalka during the Spring '08 term at UC Davis.

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16A-MT2Solution - 1 Mat16A-MT2-5/14/10 1 2 3 TOT 4+4 pt 4+4...

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