The derivative of $\frac1x$ is $\frac{-1}{x^2}$.
How do I find the $c$ if there is no zero in the derivative of the function?
I started with $-1/x^2= -0.0625$ but I'm confused from here on.
$\endgroup$ 12 Answers
$\begingroup$The question is asking you to solve for
$$f'(c) = -0.0625$$
where $$c \in (2,8)$$
Hence, we need to solve for $$-\frac{1}{c^2}=-0.0625$$
$$c^2=\frac{1}{0.0625}$$
Solve for $c \in (2,8)$.
$\endgroup$ $\begingroup$$$f (x)=\frac {1}{x} $$ $f $ is continuous at $[2,8] $ and differentiable at $(2,8) $ thus by MVT, $$\exists c\in (2,8) : $$ $$f (8)-f (2)=(8-2)f'(c)$$ $$\frac {1}{8}-\frac {1}{2}=-6.\frac {1}{c^2} $$
$$\frac {-6}{16}=\frac {-6}{c^2} $$ $$\implies c=4$$
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