Q. 91 - Calculus

Question

Prove part (b) of Theorem $3.8$: With hypotheses as stated in the theorem, if $x=c$ is a critical point of f, where $f^{\text{'}}\left(x\right)<0$ to the left of c and $f^{\text{'}}\left(x\right)>0$ to the right of c, then f has a local minimum at $x=c$.

Step-by-Step Solution

Answer

The part(b) of theorem $3.8$ is proved.

Step 1. Given Information

We are given a function f and theorem $3.8$.

Step 2. Proving the statement

Suppose $f^{\text{'}}\left(x\right)<0$ for $x\in (a,c)$ and $f^{\text{'}}\left(x\right)>0$ for $x\in (c,b)$, that is suppose f is decreasing on $(a, c]$ and increasing on $[c, b)$. We will show that $f\left(c\right)\le f\left(x\right)$ for all $x\in (a,b)$ which will tell us that f has local minimum at $x=c$.

Given that $x\in (a,b)$, there will be $3$ cases to consider.

First if $x=c$ then clearly $f\left(c\right)=f\left(x\right)$.

Second if $a<x<c$ then since f is decreasing on $(a, c]$ we have $f\left(x\right)>f\left(c\right)$.

Third if $c<x<b$ then since f is increasing on $[c, b)$, we have $f\left(x\right)>f\left(c\right)$.

In all the three cases we have $f\left(x\right)\ge f\left(c\right)$ and therefore f has a local minimum at $x=c$.