What is the extreme f(x)=16x^4-96x^2 ?

The extreme f(x)=16x^4-96x^2 is Minimum(-sqrt(3),-144),Maximum(0,0),Minimum(sqrt(3),-144)

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extreme f(x)=16x^4-96x^2

extreme points

f (x) = 16 x^{4} - 96 x^{2}

Minimum (- 3, - 144), Maximum (0, 0), Minimum (3, - 144)

Solution steps

f^{'} (x) = 64 x^{3} - 192 x

Find intervals:

Decreasing

: - \infty < x < - 3,

Increasing

: - 3 < x < 0,

Decreasing

: 0 < x < 3,

Increasing

: 3 < x < \infty

Plug

x = - 3

into

16 x^{4} - 96 x^{2} : - 144

Minimum (- 3, - 144)

Plug

x = 0

into

16 x^{4} - 96 x^{2} : 0

Maximum (0, 0)

Plug

x = 3

into

16 x^{4} - 96 x^{2} : - 144

Minimum (3, - 144)

Minimum (- 3, - 144), Maximum (0, 0), Minimum (3, - 144)

(t,) (,)

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

f (x) = x - 5

f (x) = - \frac{2}{3} x + 2

What is the extreme f(x)=16x^4-96x^2 ?
The extreme f(x)=16x^4-96x^2 is Minimum(-sqrt(3),-144),Maximum(0,0),Minimum(sqrt(3),-144)