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

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Answer 1 · 2015-09-26T18:42:33

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

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

How do use the first derivative test to determine the local extrema 36x^2 +24x^236x2+24x236x^2 +24x^2?