What is the extreme f(x)= 1/2 x^4-4x^2+3 ?

The extreme f(x)= 1/2 x^4-4x^2+3 is Minimum(-2,-5),Maximum(0,3),Minimum(2,-5)

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extreme f(x)= 1/2 x^4-4x^2+3

extreme points

f (x) = \frac{1}{2} x^{4} - 4 x^{2} + 3

Minimum (- 2, - 5), Maximum (0, 3), Minimum (2, - 5)

Solution steps

f^{'} (x) = 2 x^{3} - 8 x

Find intervals:

Decreasing

: - \infty < x < - 2,

Increasing

: - 2 < x < 0,

Decreasing

: 0 < x < 2,

Increasing

: 2 < x < \infty

Plug

x = - 2

into

\frac{1}{2} x^{4} - 4 x^{2} + 3 : - 5

Minimum (- 2, - 5)

Plug

x = 0

into

\frac{1}{2} x^{4} - 4 x^{2} + 3 : 3

Maximum (0, 3)

Plug

x = 2

into

\frac{1}{2} x^{4} - 4 x^{2} + 3 : - 5

Minimum (2, - 5)

Minimum (- 2, - 5), Maximum (0, 3), Minimum (2, - 5)

f (x) = - 3 x^{2} + 180 x + 190, 0 \leq x \leq 40

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

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

\frac{x ^{2} - 1}{x ^{2} + 4}

f (x, y) = y - x

What is the extreme f(x)= 1/2 x^4-4x^2+3 ?
The extreme f(x)= 1/2 x^4-4x^2+3 is Minimum(-2,-5),Maximum(0,3),Minimum(2,-5)