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Question
Hello,
I was wondering if you could show me how to solve the following question:

The function  on [-8,8] does not satisfy the conditions of the mean-value theorem because

A. f(0) is not defined.   
B. f(x) is not continuous on [-8,8].     
C. f (-1) does not exist.         
D. f(x) is not defined for x < 0.
E. f (0) does not exist.

Thank you in advance for all of your help!

Answer
A B C D E.

Ex. Let f(x) = xsin(1/x) x not = 0.
         = 0         x = 0.

f'(x) = sin(1/x) - 1/x(cos(1/x)   xn0t = 0.

At x = 0 the mean val. thm.does not apply, since 1/x is not defined at x = 0

For t not = 0

[f(t) - f(0)]/(t - 0) = sin(1/t).

As t tends to 0, this does not tend to any limit, so that f'(0) does not exist.

Calculus

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mervyn

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I can solve problems in calculus, advanced calculus, differential and partial differential equations, complex variables, Fourier series and integrals, analysis,Gram-Schmidt orthogonalization, and Legendre polynomials. Soon I will be able to answer any question on advanced engineering control.

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