What is the extreme f(x)=x^4-32x^2+3 ?

The extreme f(x)=x^4-32x^2+3 is Minimum(-4,-253),Maximum(0,3),Minimum(4,-253)

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extreme f(x)=x^4-32x^2+3

extreme points

f (x) = x^{4} - 32 x^{2} + 3

Minimum (- 4, - 253), Maximum (0, 3), Minimum (4, - 253)

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} + 3 : - 253

Minimum (- 4, - 253)

Plug

x = 0

into

x^{4} - 32 x^{2} + 3 : 3

Maximum (0, 3)

Plug

x = 4

into

x^{4} - 32 x^{2} + 3 : - 253

Minimum (4, - 253)

Minimum (- 4, - 253), Maximum (0, 3), Minimum (4, - 253)

x^{4} + 2 x^{3} + x^{2} + 2 x + 2

y = 3 x + 2 cos (x)

f (x, y) = x^{3} + y^{2} - 6 x y - 9 x + 8 y + 2

y = sin (x) + cos (x)

f (x) = 4 x^{2} + 40, - 1 \leq x \leq 1

What is the extreme f(x)=x^4-32x^2+3 ?
The extreme f(x)=x^4-32x^2+3 is Minimum(-4,-253),Maximum(0,3),Minimum(4,-253)