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

Sep 26, 2015

$y = 36 {x}^{2} + 24 {x}^{2} = 60 {x}^{2}$

#### Explanation:

$y ' = 120 x$ 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 $x = 0$).