What is the extreme f(x)=x^4-8x^2+5 ?

The extreme f(x)=x^4-8x^2+5 is Minimum(-2,-11),Maximum(0,5),Minimum(2,-11)

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extreme f(x)=x^4-8x^2+5

extreme points

f (x) = x^{4} - 8 x^{2} + 5

Minimum (- 2, - 11), Maximum (0, 5), Minimum (2, - 11)

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} + 5 : - 11

Minimum (- 2, - 11)

Plug

x = 0

into

x^{4} - 8 x^{2} + 5 : 5

Maximum (0, 5)

Plug

x = 2

into

x^{4} - 8 x^{2} + 5 : - 11

Minimum (2, - 11)

Minimum (- 2, - 11), Maximum (0, 5), Minimum (2, - 11)

f (x) = 4 x^{3} - 12 x^{2}

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

f (x) = x^{3} + x^{2} - x

f (x, y) = x - 9 y + 17

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

What is the extreme f(x)=x^4-8x^2+5 ?
The extreme f(x)=x^4-8x^2+5 is Minimum(-2,-11),Maximum(0,5),Minimum(2,-11)