What is the extreme f(x)=8x^4-48x^2 ?

The extreme f(x)=8x^4-48x^2 is Minimum(-sqrt(3),-72),Maximum(0,0),Minimum(sqrt(3),-72)

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extreme f(x)=8x^4-48x^2

extreme points

f (x) = 8 x^{4} - 48 x^{2}

Minimum (- 3, - 72), Maximum (0, 0), Minimum (3, - 72)

Solution steps

f^{'} (x) = 32 x^{3} - 96 x

Find intervals:

Decreasing

: - \infty < x < - 3,

Increasing

: - 3 < x < 0,

Decreasing

: 0 < x < 3,

Increasing

: 3 < x < \infty

Plug

x = - 3

into

8 x^{4} - 48 x^{2} : - 72

Minimum (- 3, - 72)

Plug

x = 0

into

8 x^{4} - 48 x^{2} : 0

Maximum (0, 0)

Plug

x = 3

into

8 x^{4} - 48 x^{2} : - 72

Minimum (3, - 72)

Minimum (- 3, - 72), Maximum (0, 0), Minimum (3, - 72)

f (x) = \frac{1}{x + 1}

r (x) = \frac{6 x ^{3} - 2}{2 x ^{3} + 5 x ^{2} + 6 x}

(3, - 2), (- 4, 5)

f (x) = 5 x^{2} + 9

f (x) = \frac{x - 1}{x - 4}

What is the extreme f(x)=8x^4-48x^2 ?
The extreme f(x)=8x^4-48x^2 is Minimum(-sqrt(3),-72),Maximum(0,0),Minimum(sqrt(3),-72)