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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!

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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