Answer
$c=\frac{1}{3}$ is the only critical number of $f$.
Work Step by Step
How to find the critical numbers of a function according to definition
1) Find all numbers of $c$ satisfying either $f'(c)=0$ or $f'(c)$ does not exist.
2) See that all the $c$ we get lie in the domain of $f$ $(D_f)$ or not.
- If $c$ lies in $D_f$, $c$ is a critical number of $f$.
- If $c$ does not lie in $D_f$, $c$ is not a critical number of $f$.
$$f(x)=4+\frac{1}{3}x-\frac{1}{2}x^2$$
$D_f=R$
First, we find $f'(x)$ $$f'(x)=0+\frac{1}{3}\times1-\frac{1}{2}\times2x$$ $$f'(x)=\frac{1}{3}-x$$
1) Find $f'(c)=0$ $$f'(c)=0$$ $$\frac{1}{3}-c=0$$ $$c=\frac{1}{3}$$
There are no instances of $c$ where $f'(c)$ does not exist in this case.
2) Examine whether $c$ lies in $D_f$ or not.
We see that $c=\frac{1}{3}\in R$, so $c=\frac{1}{3}$ lies in $D_f$.
We conclude that $c=\frac{1}{3}$ is a critical number of $f$.
