What is the extreme f(x)=(x^4-32x^2)/9 ?

The extreme f(x)=(x^4-32x^2)/9 is Minimum(-4,-256/9),Maximum(0,0),Minimum(4,-256/9)

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extreme f(x)=(x^4-32x^2)/9

extreme

Minimum (- 4, - \frac{256}{9}), Maximum (0, 0), Minimum (4, - \frac{256}{9})

Solution steps

f^{'} (x) = \frac{1}{9} (4 x^{3} - 64 x)

Find intervals:

Decreasing

: - \infty < x < - 4,

Increasing

: - 4 < x < 0,

Decreasing

: 0 < x < 4,

Increasing

: 4 < x < \infty

Plug

x = - 4

into

\frac{x ^{4} - 32 x ^{2}}{9} : - \frac{256}{9}

Minimum (- 4, - \frac{256}{9})

Plug

x = 0

into

\frac{x ^{4} - 32 x ^{2}}{9} : 0

Maximum (0, 0)

Plug

x = 4

into

\frac{x ^{4} - 32 x ^{2}}{9} : - \frac{256}{9}

Minimum (4, - \frac{256}{9})

Minimum (- 4, - \frac{256}{9}), Maximum (0, 0), Minimum (4, - \frac{256}{9})

What is the extreme f(x)=(x^4-32x^2)/9 ?
The extreme f(x)=(x^4-32x^2)/9 is Minimum(-4,-256/9),Maximum(0,0),Minimum(4,-256/9)