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.

Short answer

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

Step-by-step solution

Step 1. Given Information 

We are given a function fand 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 $x=c$.