What is the extreme f(x)=x^4-8x^2+16 ?

The extreme f(x)=x^4-8x^2+16 is Minimum(-2,0),Maximum(0,16),Minimum(2,0)

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extreme f(x)=x^4-8x^2+16

extreme points

f (x) = x^{4} - 8 x^{2} + 16

Minimum (- 2, 0), Maximum (0, 16), Minimum (2, 0)

Solution steps

f^{'} (x) = 4 x^{3} - 16 x

Find intervals:

Decreasing

: - \infty < x < - 2,

Increasing

: - 2 < x < 0,

Decreasing

: 0 < x < 2,

Increasing

: 2 < x < \infty

Plug

x = - 2

into

x^{4} - 8 x^{2} + 16 : 0

Minimum (- 2, 0)

Plug

x = 0

into

x^{4} - 8 x^{2} + 16 : 16

Maximum (0, 16)

Plug

x = 2

into

x^{4} - 8 x^{2} + 16 : 0

Minimum (2, 0)

Minimum (- 2, 0), Maximum (0, 16), Minimum (2, 0)

f (x) = \frac{1}{x} - 3

\frac{x - 1}{x ^{2} + 3}

f (x) = \frac{2 x - 1}{x + 3}

f (x) = \frac{x + 6}{2 x - 4}

What is the extreme f(x)=x^4-8x^2+16 ?
The extreme f(x)=x^4-8x^2+16 is Minimum(-2,0),Maximum(0,16),Minimum(2,0)