Question

Prove the part of Theorem 3.3 that was not proved in the reading: If a function f has a local minimum at $x=c$, then either $f\left(c\right)$does not exist or $f\left(c\right)=0$.

Short answer

The part that is not proved is proved by Theorem 3.3.

Step-by-step solution

Step 1. Given Information.

The function is$f\left(x\right)$.

Step 2. Local Minimum.

Let $x=c$be the location of the local minimum of $f$.

If $f\text{'}\left(c\right)$does not exist, then $x=c$is a critical point.

We assume $f\text{'}\left(c\right)$exists, then show that $f\text{'}\left(c\right)=0$.

As $x=c$is the location of local minimum of $f$, then there exists some $\delta >0$such that,

$f\left(c\right)\le f\left(x\right)$for all $x\in (c-\delta ,c+\delta )$

That is, $f\left(x\right)-f\left(c\right)\ge 0$.

If $f\text{'}\left(c\right)$exists such that $f\text{'}\left(c\right)\ge 0$and $f\text{'}\left(c\right)\le 0$, then $f\text{'}\left(c\right)$must be equal to zero.

That is, $f\text{'}\left(c\right)=0$.

Hence, it is a function$f$has local minimum at$x=c$, then either$f\text{'}\left(c\right)$does not exist or$f\text{'}\left(c\right)=0$.