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 points

f (x) = \frac{x ^{4} - 32 x ^{2}}{9}

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})

f (x, y) = x^{3} - 3 x + x y^{2}

f (x) = [0.2 π]

(9 - x) (5 - x) (x + 9) (x + 5)

f (x) = 3 x - \frac{6}{x}, 5 \leq x \leq 7

f (x) = (x^{2} - \frac{7}{3} x + \frac{13}{9}) \cdot e^{3 x}

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)