Question

How do use the first derivative test to determine the local extrema `36x^2 +24x^2`?

Answer 1

`y=36x^2+24x^2 = 60x^2`

Explanation:

`y' = 120x` so the only critical number is `0`.

`y'` is negative left of `0` and positive right or `0`,

so the first derivative test tells us that the `y` value at `x=0` is a local minimum.

`0` is a local minimum (at `x=0`).