What is the extreme 4x^2-8x^4 ?

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

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extreme 4x^2-8x^4

extreme points

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

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

Solution steps

f^{'} (x) = 8 x - 32 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

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

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

Plug

x = 0

into

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

Minimum (0, 0)

Plug

x = \frac{1}{2}

into

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

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

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

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

f (x) = 9 x^{3} + 21 x^{2} + 8 x - 3

f (x) = x^{2} - 2 x + 5 [- 1.4]

2 x^{4} - 196 x^{2} - 3

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

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