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Let $f(x) = \frac{1}{x - 2}$ on the closed interval [3, 6].
Part a) Find the average rate of change for $f$ on [3, 6]. (No decimal answers. If necessary, enter your
answer as a reduced fraction.)
Average rate of change =
1
Part b) The numerator and denominator of $f(x) = \frac{1}{x - 2}$ are both polynomials, and polynomials are
continuous and differentiable everywhere. So $f$ is continuous when $x - 2 \neq 0$. And, since 2 is not in the
closed interval [3, 6], $f$ is continuous on [3, 6] and differentiable on (3, 6).
This means the $f$ satisfies the conditions of the Mean Value Theorem.
Apply the Mean Value Theorem to find all numbers $c$ in (3, 6) where $f'(c)$ is equal to the average rate of
change. (If more than one value of $c$ exists, enter your answers as a comma separated list. If $c$ does not
exist, enter DNE.)
c =
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