What is the extreme f(x)=2x^2-8x^4 ?

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

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extreme f(x)=2x^2-8x^4

extreme

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

Solution steps

f^{'} (x) = 4 x - 32 x^{3}

Find intervals:

Increasing

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

Decreasing

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

Increasing

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

Decreasing

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

Plug

x = - \frac{1}{2 2}

into

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

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

Plug

x = 0

into

2 x^{2} - 8 x^{4} : 0

Minimum (0, 0)

Plug

x = \frac{1}{2 2}

into

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

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

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

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