Q. 92 - Calculus

Question

Prove part (d) 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 and to the right of c, then $x=c$ is not a local extremum of f.

Step-by-Step Solution

Answer

The part (d) of Theorem $3.8$ is proved.

Step 1. Given Information

We are given a function f and part (d) of Theorem $3.8$.

Step 2. Proving the statement

Let $f^{\text{'}}\left(x\right)<0$ for all $x\in (a,c)\cup (c,b)$. Then f must be decreasing on all of $(a, b)$.

Now the point $x=c$ cannot be the location of a local minimum of f because for all $x<c$ in $(a, b), f\left(x\right)<f\left(c\right)$. But neither $x=c$ cannot be the location of a local maximum because for all $x>c$  in $(a, b)$,

$f\left(x\right)>f\left(c\right)$

Since f is increasing on $(a, b)$.

Therefore f has neither a local minimum nor a local maximum at $x=c$.