Answer
Critical number: $t=\frac{4}{3}$.
Work Step by Step
The given function is:
$g(t)=|3t-4|$
When $3t-4 \lt 0$ or $t\lt\frac{4}{3}$ then $g(t)=-(3t-4)=-3t+4$
When $3t-4 \gt 0$ or $t\gt \frac{4}{3}$ then $g(t)=3t-4$
When $3t-4 = 0$ or $t=\frac{4}{3}$ then $g(t)=0$
Notice that all the linear functions are differentiable on $\mathbb R$.
The critical number $c$ of $g$ should satisfy the condition $g'(c)=0$.
So when $3t-4 \lt 0$ or $t\lt\frac{4}{3}$ the first derivative of $g$ with respect to $t$ is $-3$.
So when $3t-4 \gt 0$ or $t\gt\frac{4}{3}$ the first derivative of $g$ with respect to $t$ is $3$.
When $3t-4 = 0$ or $t=\frac{4}{3}$ then $g'(t)=0$
Since $g'(t)=0$ for $t=\frac{4}{3}$ it follows that $g$ have critical number at $t=\frac{4}{3}$.