What is the extreme y=x^2-x^4 ?

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

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extreme y=x^2-x^4

extreme points

y = x^{2} - x^{4}

Maximum (- \frac{1}{2}, \frac{1}{4}), Minimum (0, 0), Maximum (\frac{1}{2}, \frac{1}{4})

Solution steps

f^{'} (x) = 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

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

Maximum (- \frac{1}{2}, \frac{1}{4})

Plug

x = 0

into

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

Minimum (0, 0)

Plug

x = \frac{1}{2}

into

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

Maximum (\frac{1}{2}, \frac{1}{4})

Maximum (- \frac{1}{2}, \frac{1}{4}), Minimum (0, 0), Maximum (\frac{1}{2}, \frac{1}{4})

sin (2 x + 1)

f (x) = (2 x) - \frac{3}{5}

f (x) = 5 + x^{2} - 4 x

\frac{2 x}{3 x - 8}

f (x) = e^{x} (x^{2} + x) x - sin (x)

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