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

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Answer 1 · 2016-11-04T17:19:38

See below.

$g(t) = abs(3t-4) = { (3t-4,"if",t >= 4/3),(-3t+4,"if",t < 4/3 ) :}$

$g'(t) = { (3,"if",t >= 4/3),(-3,"if",t < 4/3 ) :}$

$g'$ is never $0$ and is undefined (fals to exist) at $x= 4/3$

The only critical number is $4/3$ .

We see that $g$ is decreasing left of $4/3$ and increasing on the right.

So $g(4/3) = 0$ is a local minimum.

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