Q.64 - Calculus

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 $f\left(c\right)=0$.

Step-by-Step Solution

Answer

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

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 $f\text{'}\left(c\right)=0$.