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

The extreme f(x)=(x^4)/4-(3x^2)/2 is Minimum(-sqrt(3),-9/4),Maximum(0,0),Minimum(sqrt(3),-9/4)

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

extreme

Minimum (- 3, - \frac{9}{4}), Maximum (0, 0), Minimum (3, - \frac{9}{4})

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < - 3,

Increasing

: - 3 < x < 0,

Decreasing

: 0 < x < 3,

Increasing

: 3 < x < \infty

Plug

x = - 3

into

\frac{x ^{4}}{4} - \frac{3 x ^{2}}{2} : - \frac{9}{4}

Minimum (- 3, - \frac{9}{4})

Plug

x = 0

into

\frac{x ^{4}}{4} - \frac{3 x ^{2}}{2} : 0

Maximum (0, 0)

Plug

x = 3

into

\frac{x ^{4}}{4} - \frac{3 x ^{2}}{2} : - \frac{9}{4}

Minimum (3, - \frac{9}{4})

Minimum (- 3, - \frac{9}{4}), Maximum (0, 0), Minimum (3, - \frac{9}{4})

What is the extreme f(x)=(x^4)/4-(3x^2)/2 ?
The extreme f(x)=(x^4)/4-(3x^2)/2 is Minimum(-sqrt(3),-9/4),Maximum(0,0),Minimum(sqrt(3),-9/4)