Question #6bad8

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Answer 1 · 2017-02-20T13:34:47

For bounds on $f(5)$ , please see below.

We are told that $f$ is differentiable.

We are not given the domain of $f$ , but in order to apply the following, the domain must include $[1,5]$

Since $f$ is (we assume) differentiable on $[1,5]$ and continuous on $(1,5)$ , we can apply the Mean Value Theorem to conclude that

for some $c$ in $(1,5)$ , we have

$f(5) - f(1) = f'(c)(5-1)$ . That is

$f(5) = 4f'(c) + pi$ for some $c$ in $(1,5)$

We are also told that $f'(x) = sqrt(x^3+6)$ .

Since the derivative of $sqrt(x^3+6)$ is positive, $f'$ is an increasing function.

Therefore, for the $c$ mentioned above, we have

$f'(1) < f'(c) < f'(5)$

$sqrt7 < f'(c) < sqrt131$ .

We conclude that

$4sqrt7+pi < f(5) < 4sqrt131+pi$