What is the extreme f(x)=x^4-72x^2+9 ?

The extreme f(x)=x^4-72x^2+9 is Minimum(-6,-1287),Maximum(0,9),Minimum(6,-1287)

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extreme f(x)=x^4-72x^2+9

extreme points

f (x) = x^{4} - 72 x^{2} + 9

Minimum (- 6, - 1287), Maximum (0, 9), Minimum (6, - 1287)

Solution steps

f^{'} (x) = 4 x^{3} - 144 x

Find intervals:

Decreasing

: - \infty < x < - 6,

Increasing

: - 6 < x < 0,

Decreasing

: 0 < x < 6,

Increasing

: 6 < x < \infty

Plug

x = - 6

into

x^{4} - 72 x^{2} + 9 : - 1287

Minimum (- 6, - 1287)

Plug

x = 0

into

x^{4} - 72 x^{2} + 9 : 9

Maximum (0, 9)

Plug

x = 6

into

x^{4} - 72 x^{2} + 9 : - 1287

Minimum (6, - 1287)

Minimum (- 6, - 1287), Maximum (0, 9), Minimum (6, - 1287)

\frac{4}{x ^{2} + x - 2}

f (x) = \frac{1 + x ^{2}}{4}

\frac{x}{x + 1}

(6, 3), (10, - 7)

f (x) = \frac{3 x + 5}{7 x + 4}

What is the extreme f(x)=x^4-72x^2+9 ?
The extreme f(x)=x^4-72x^2+9 is Minimum(-6,-1287),Maximum(0,9),Minimum(6,-1287)