Question

Question #6bad8

Answer 1

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

Explanation:

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`