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

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

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

extreme points

f (x) = x^{4} - x^{2}

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

Solution steps

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

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

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

Plug

x = 0

into

x^{4} - x^{2} : 0

Maximum (0, 0)

Plug

x = \frac{1}{2}

into

x^{4} - x^{2} : - \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})

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

f (x, y) = 2 x^{2} + 4 x y + 6 y^{2} - 8 x - 10

f (x, y) = x e^{- 2 x^{2} - 2 y^{2}}

4 x y

f (x) = \frac{- 2 x ^{2} + 5 x - 1}{2 x - 1}

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