What is the extreme y=(2x^2-1)^2 ?

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

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extreme y=(2x^2-1)^2

extreme points

y = (2 x^{2} - 1)^{2}

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

Solution steps

f^{'} (x) = 8 x (2 x^{2} - 1)

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^{2} - 1)^{2} : 0

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

Plug

x = 0

into

(2 x^{2} - 1)^{2} : 1

Maximum (0, 1)

Plug

x = \frac{1}{2}

into

(2 x^{2} - 1)^{2} : 0

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

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

f (x) = \frac{6}{- 4 x + 2}

f (y) = x^{y + 1}

f (x) = x^{2} lo g_{8} (x)

x + \frac{192}{x}

f (x) = - \frac{x ^{2}}{10} + 2 x + 3

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