Question

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

Answer 1

See below.

Explanation:

`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.