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

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

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

extreme points

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

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

Solution steps

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

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

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

Plug

x = 0

into

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

Minimum (0, 0)

Plug

x = \frac{1}{2}

into

2 x^{2} (1 - x^{2}) : \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) = x^{2} + y

f (x, y) = 12 x y - x^{3} - 6 y^{2}

x^{2} + 2 x

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

f (x) = (x - 4)^{2} + 1 + 3

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