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

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

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

extreme points

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

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

Solution steps

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

Find intervals:

Increasing

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

Decreasing

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

Increasing

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

Decreasing

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

Plug

x = - \frac{1}{2}

into

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

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

Plug

x = 0

into

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

Minimum (0, 0)

Plug

x = \frac{1}{2}

into

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

Maximum (\frac{1}{2}, \frac{1}{8})

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

f (x) = (x - 8) e^{x}

y = 6 t^{2} - 42 t - 60

y = y = \frac{x ^{3}}{3} + \frac{x ^{2}}{2} - 2 x + 9

- y^{2} + x y - x^{3} + y + 3

f (x, y) = - x^{4} + 8 x^{2} - y^{2} - 1

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