# How do you find the critical numbers for g(t)=abs(3t-4) to determine the maximum and minimum?

Nov 4, 2016

See below.

#### Explanation:

$g \left(t\right) = \left\mid 3 t - 4 \right\mid = \left\{\begin{matrix}3 t - 4 & \text{if" & t >= 4/3 \\ -3t+4 & "if} & t < \frac{4}{3}\end{matrix}\right.$

$g ' \left(t\right) = \left\{\begin{matrix}3 & \text{if" & t >= 4/3 \\ -3 & "if} & t < \frac{4}{3}\end{matrix}\right.$

$g '$ is never $0$ and is undefined (fals to exist) at $x = \frac{4}{3}$

The only critical number is $\frac{4}{3}$.

We see that $g$ is decreasing left of $\frac{4}{3}$ and increasing on the right.

So $g \left(\frac{4}{3}\right) = 0$ is a local minimum.