What is the extreme f(x)=x^4-288x^2+20736 ?

The extreme f(x)=x^4-288x^2+20736 is Minimum(-12,0),Maximum(0,20736),Minimum(12,0)

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extreme f(x)=x^4-288x^2+20736

extreme points

f (x) = x^{4} - 288 x^{2} + 20736

Minimum (- 12, 0), Maximum (0, 20736), Minimum (12, 0)

Solution steps

f^{'} (x) = 4 x^{3} - 576 x

Find intervals:

Decreasing

: - \infty < x < - 12,

Increasing

: - 12 < x < 0,

Decreasing

: 0 < x < 12,

Increasing

: 12 < x < \infty

Plug

x = - 12

into

x^{4} - 288 x^{2} + 20736 : 0

Minimum (- 12, 0)

Plug

x = 0

into

x^{4} - 288 x^{2} + 20736 : 20736

Maximum (0, 20736)

Plug

x = 12

into

x^{4} - 288 x^{2} + 20736 : 0

Minimum (12, 0)

Minimum (- 12, 0), Maximum (0, 20736), Minimum (12, 0)

(0, s), (s, 0)

f (x) = 4^{x}

\frac{2 x - 2}{x + 2}

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

f (x) = (5 - x)^{\frac{1}{4}}

What is the extreme f(x)=x^4-288x^2+20736 ?
The extreme f(x)=x^4-288x^2+20736 is Minimum(-12,0),Maximum(0,20736),Minimum(12,0)