What is the extreme x^4-1/2 x^2-2 ?

The extreme x^4-1/2 x^2-2 is Minimum(-1/2 ,-33/16),Maximum(0,-2),Minimum(1/2 ,-1/16-2)

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extreme x^4-1/2 x^2-2

extreme points

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

Minimum (- \frac{1}{2}, - \frac{33}{16}), Maximum (0, - 2), Minimum (\frac{1}{2}, - \frac{1}{16} - 2)

Solution steps

f^{'} (x) = 4 x^{3} - x

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

x^{4} - \frac{1}{2} x^{2} - 2 : - \frac{33}{16}

Minimum (- \frac{1}{2}, - \frac{33}{16})

Plug

x = 0

into

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

Maximum (0, - 2)

Plug

x = \frac{1}{2}

into

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

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

Minimum (- \frac{1}{2}, - \frac{33}{16}), Maximum (0, - 2), Minimum (\frac{1}{2}, - \frac{1}{16} - 2)

3 x^{5} - 5 x^{3} + 2

y = e^{- 4 t} + e^{- 5 t}

f (x) = \frac{2 x}{x ^{4} + 1}

f (x) = 3 x - ln (3 x), (0, 4)

2 x^{3} - 9 x

What is the extreme x^4-1/2 x^2-2 ?
The extreme x^4-1/2 x^2-2 is Minimum(-1/2 ,-33/16),Maximum(0,-2),Minimum(1/2 ,-1/16-2)