What is the extreme f(x)=x^4-32x^2+7 ?

The extreme f(x)=x^4-32x^2+7 is Minimum(-4,-249),Maximum(0,7),Minimum(4,-249)

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extreme f(x)=x^4-32x^2+7

extreme points

f (x) = x^{4} - 32 x^{2} + 7

Minimum (- 4, - 249), Maximum (0, 7), Minimum (4, - 249)

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} + 7 : - 249

Minimum (- 4, - 249)

Plug

x = 0

into

x^{4} - 32 x^{2} + 7 : 7

Maximum (0, 7)

Plug

x = 4

into

x^{4} - 32 x^{2} + 7 : - 249

Minimum (4, - 249)

Minimum (- 4, - 249), Maximum (0, 7), Minimum (4, - 249)

f (x, y) = y^{3} + 3 x^{2} - 12 x - 6 y^{2}

f (x, y) = 2^{2} x^{4} + 2^{2} y^{4} - 4 x y + 13

f (x) = x^{7} e^{- x}, - 4 \leq x \leq 11

f (x, y) = x y - x - 2 y + 1

f (x) = ln (1 - x^{2} - y^{2})

What is the extreme f(x)=x^4-32x^2+7 ?
The extreme f(x)=x^4-32x^2+7 is Minimum(-4,-249),Maximum(0,7),Minimum(4,-249)