Question

Suppose f is a differentiable function such that f′(x)≤2 for all x∈[1,3]. If f(1)=4, the Mean Value Theorem says that f(3)≤V for what value of V? (Choose V as small as possible.)

Answer 1

The MVT guarantees the existence of $c\in(1,3)$ such that

$f(c)=\dfrac{f(3)-f(1)}{3-1}=\dfrac{f(3)-4}2$

Since $f'(x)\le2$ for all $x\in[1,3]$, we have

$\dfrac{f(3)-4}2\le2\implies f(3)-4\le4\implies f(3)\le8$

so that $V=8$.