What is the extreme f(x)=x^4-32x^2+8 ?

The extreme f(x)=x^4-32x^2+8 is Minimum(-4,-248),Maximum(0,8),Minimum(4,-248)

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extreme f(x)=x^4-32x^2+8

extreme points

f (x) = x^{4} - 32 x^{2} + 8

Minimum (- 4, - 248), Maximum (0, 8), Minimum (4, - 248)

Solution steps

f^{'} (x) = 4 x^{3} - 64 x

Find intervals:

Decreasing

: - \infty < x < - 4,

Increasing

: - 4 < x < 0,

Decreasing

: 0 < x < 4,

Increasing

: 4 < x < \infty

Plug

x = - 4

into

x^{4} - 32 x^{2} + 8 : - 248

Minimum (- 4, - 248)

Plug

x = 0

into

x^{4} - 32 x^{2} + 8 : 8

Maximum (0, 8)

Plug

x = 4

into

x^{4} - 32 x^{2} + 8 : - 248

Minimum (4, - 248)

Minimum (- 4, - 248), Maximum (0, 8), Minimum (4, - 248)

f (x, y) = 2 x^{3} - 5 x y^{2} + 4 y^{3}

f (x, y) = 2 x^{2} - x y + y^{2} - 6 x - 9 y + 7

12 x^{\frac{2}{3}} - 8 x

f (x) = 9 - x^{2}, - 3 \leq x \leq 2

f (x) = \frac{1}{x} - \frac{64}{y} + x y

What is the extreme f(x)=x^4-32x^2+8 ?
The extreme f(x)=x^4-32x^2+8 is Minimum(-4,-248),Maximum(0,8),Minimum(4,-248)