What are the local extrema, if any, of $f (x) = 120 x^{5} - 200 x^{3}$ $f(x)= 120x^5 - 200x^3$ ?

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Answer 1 · 2016-03-11T14:49:53

Local maximum of $80$ $80$ (at $x = - 1$ $x=-1$ ) and local minimum of $- 80$ $-80$ (at $x = 1$ $x=1$ .

$f (x) = 120 x^{5} - 200 x^{3}$ $f(x)= 120x^5 - 200x^3$

$f'(x)= 600x^4 - 600x^2 = 600x^2(x^2 - 1)$

Critical numbers are: $-1$ , $0$ , and $1$

The sign of $f'$ changes from + to - as we pass $x=-1$ , so $f(-1) = 80$ is a local maximum.

(Since $f$ is odd, we can immediately conclude that $f(1)=-80$ is a relative minimum and $f(0)$ is not a local extremum.)

The sign of $f'$ does not change as we pass $x= 0$ , so $f(0)$ is not a local extremum.

The sign of $f'$ changes from - to + as we pass $x=1$ , so $f(1) = -80$ is a local minimum.

What are the local extrema, if any, of f(x)= 120x^5 - 200x^3f(x)=120x5−200x3f(x)= 120x^5 - 200x^3?