Answer
$c=\frac{4}{3}$ is the only critical number of $g$.
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 $f$.
$$g(t)=|3t-4|$$
$D_g=R$
First, we need to deal with the absolute value.
$$t\geq\frac{4}{3}\hspace{0.5cm}then\hspace{0.5cm}|3t-4|=3t-4 \\ t\lt\frac{4}{3}\hspace{0.5cm}then\hspace{0.5cm}|3t-4|=4-3t$$
1) Now, we find $g'(t)$
- For $t\gt\frac{4}{3}$, $$g'(t)=(3t-4)'=3$$
- For $t\lt\frac{4}{3}$, $$g'(t)=(4-3t)'=-3$$
We see that there are no instances of $c$ where $g'(c)=0$ in this case.
However, since there are 2 different values of $g'(t)$ for $t\gt\frac{4}{3}$ and $t\lt\frac{4}{3}$, $g'(\frac{4}{3})$ does not exist.
2) Examine whether $c$ lies in $D_g$ or not.
We see that $c=\frac{4}{3}\in R$, so $c=\frac{4}{3}$ lies in $D_g$.
We conclude that $c=\frac{4}{3}$ is a critical number of $g$.
