What is the extreme f(x)=x^2(x^2-1) ?

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

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extreme f(x)=x^2(x^2-1)

extreme

Minimum (- \frac{1}{2}, - \frac{1}{4}), Maximum (0, 0), Minimum (\frac{1}{2}, - \frac{1}{4})

Solution steps

f^{'} (x) = 2 x (2 x + 1) (2 x - 1)

Find intervals:

Decreasing

: - \infty < x < - \frac{1}{2},

Increasing

: - \frac{1}{2} < x < 0,

Decreasing

: 0 < x < \frac{1}{2},

Increasing

: \frac{1}{2} < x < \infty

Plug

x = - \frac{1}{2}

into

x^{2} (x^{2} - 1) : - \frac{1}{4}

Minimum (- \frac{1}{2}, - \frac{1}{4})

Plug

x = 0

into

x^{2} (x^{2} - 1) : 0

Maximum (0, 0)

Plug

x = \frac{1}{2}

into

x^{2} (x^{2} - 1) : - \frac{1}{4}

Minimum (\frac{1}{2}, - \frac{1}{4})

Minimum (- \frac{1}{2}, - \frac{1}{4}), Maximum (0, 0), Minimum (\frac{1}{2}, - \frac{1}{4})

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