What is the extreme-(x^4)/4+(x^2)/4 ?

The extreme-(x^4)/4+(x^2)/4 is Maximum(-sqrt(1/2), 1/16),Minimum(0,0),Maximum(sqrt(1/2), 1/16)

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extreme-(x^4)/4+(x^2)/4

extreme

Maximum (- \frac{1}{2}, \frac{1}{16}), Minimum (0, 0), Maximum (\frac{1}{2}, \frac{1}{16})

Solution steps

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

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

- \frac{x ^{4}}{4} + \frac{x ^{2}}{4} : \frac{1}{16}

Maximum (- \frac{1}{2}, \frac{1}{16})

Plug

x = 0

into

- \frac{x ^{4}}{4} + \frac{x ^{2}}{4} : 0

Minimum (0, 0)

Plug

x = \frac{1}{2}

into

- \frac{x ^{4}}{4} + \frac{x ^{2}}{4} : \frac{1}{16}

Maximum (\frac{1}{2}, \frac{1}{16})

Maximum (- \frac{1}{2}, \frac{1}{16}), Minimum (0, 0), Maximum (\frac{1}{2}, \frac{1}{16})

What is the extreme-(x^4)/4+(x^2)/4 ?
The extreme-(x^4)/4+(x^2)/4 is Maximum(-sqrt(1/2), 1/16),Minimum(0,0),Maximum(sqrt(1/2), 1/16)