Question

If $f\left(1\right)=10$ and $f^{\prime }\left(x\right)\ge 2$ for $1\le x\le 4$, how small $f\left(4\right)$ can possibly be?

Answer 1

Recall the Mean Value Theorem.

If a function is continuous on some closed interval $[a,b]$ and differentiable on the open interval $(a,b)$ then there exists a point $c$ such that $f^{\prime }\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$

In this particular question, we assume that the function is continuous because not enough information was supplied. Also note that the interval is $[1,4]$

So according to MVT, we have

$\exists cϵ(1,4)$ such that,:

$f^{\prime }\left(c\right)=\frac{f\left(4\right)-f\left(1\right)}{4-1}$

Hence $f^{\prime }\left(c\right)=\frac{f\left(4\right)-10}{3}$

Now we are given that $f^{\prime }\left(x\right)\ge 2$ for all the $xϵ[1,4]$.

Use this relation to proceed.

So, $f^{\prime }\left(c\right)\ge 2$

Hence, $2\le \frac{f\left(4\right)-10}{3}$

Thus, $f\left(4\right)\ge 16$