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

The extreme f(x)=2x^4-x^2 is Minimum(-1/2 ,-1/8),Maximum(0,0),Minimum(1/2 ,-1/8)

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

extreme points

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

Minimum (- \frac{1}{2}, - \frac{1}{8}), Maximum (0, 0), Minimum (\frac{1}{2}, - \frac{1}{8})

Solution steps

f^{'} (x) = 8 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

2 x^{4} - x^{2} : - \frac{1}{8}

Minimum (- \frac{1}{2}, - \frac{1}{8})

Plug

x = 0

into

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

Maximum (0, 0)

Plug

x = \frac{1}{2}

into

2 x^{4} - x^{2} : - \frac{1}{8}

Minimum (\frac{1}{2}, - \frac{1}{8})

Minimum (- \frac{1}{2}, - \frac{1}{8}), Maximum (0, 0), Minimum (\frac{1}{2}, - \frac{1}{8})

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

p (x) = 12 x^{4} + 15 x^{3} + a x^{2} - 69 x - 30

f (x, y) = x + 2 \frac{y}{2} x + y

f (x, y) = 16 - (x - 2)^{2} - (y - 2)^{2}

f (x) = x^{4} - 18 x^{2} - 3

What is the extreme f(x)=2x^4-x^2 ?
The extreme f(x)=2x^4-x^2 is Minimum(-1/2 ,-1/8),Maximum(0,0),Minimum(1/2 ,-1/8)