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Question #6e833

1 Answer

Apr 11, 2017

Solve the equation.

Explanation:

The conclusion of MVT says there is a $c$ $c$ in $(\frac{1}{2}, 2)$ $(1/2,2)$ such that $f'(c) = (f(2)-f(1/2))/(2-(1/2))$

Find $f'(x) = 1+1/x^2$ and $f(2)-f(1/2))/(2-(1/2)) = 2$ .

The solutions to $1+1/x^2 = 2$ are $-1$ and $1$ , but $c$ is in $(1/2,2)$ , so the only value for $c$ is $1$ .

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