Question

Suppose that 2 ≤ f '(x) ≤ 3 for all values of x. what are the minimum and maximum possible values of f(5) − f(1)?

Answer 1

Assuming $f$ is differentiable over the interval (1, 5), by the mean value theorem we know that there is some $1$ such that

$f'(c)=\dfrac{f(5)-f(1)}{5-1}$

We're given that $2\le f'(x)\le3$ for all $x$, so

$2\le\dfrac{f(5)-f(1)}4\le3\implies8\le f(5)-f(1)\le12$

that is, the minimum possible value is 8, and the maximum possible value is 12.