What is the extreme f(x)= 1/2 x^4-x^2-1 ?

The extreme f(x)= 1/2 x^4-x^2-1 is Minimum(-1,-3/2),Maximum(0,-1),Minimum(1,-3/2)

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extreme f(x)= 1/2 x^4-x^2-1

extreme points

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

Minimum (- 1, - \frac{3}{2}), Maximum (0, - 1), Minimum (1, - \frac{3}{2})

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < - 1,

Increasing

: - 1 < x < 0,

Decreasing

: 0 < x < 1,

Increasing

: 1 < x < \infty

Plug

x = - 1

into

\frac{1}{2} x^{4} - x^{2} - 1 : - \frac{3}{2}

Minimum (- 1, - \frac{3}{2})

Plug

x = 0

into

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

Maximum (0, - 1)

Plug

x = 1

into

\frac{1}{2} x^{4} - x^{2} - 1 : - \frac{3}{2}

Minimum (1, - \frac{3}{2})

Minimum (- 1, - \frac{3}{2}), Maximum (0, - 1), Minimum (1, - \frac{3}{2})

f (x) = \frac{x - 2}{x ^{2} - 6 x + 9}

f (x) = 3 + 9 x^{2} - 6 x^{3}

f (x, y) = x e^{- x^{2} - y^{2}}

f (x, y) = (a - a^{2}) x + (4 a - 2 a^{2}) y

f (x) = x (x - 4)^{\frac{1}{3}}

What is the extreme f(x)= 1/2 x^4-x^2-1 ?
The extreme f(x)= 1/2 x^4-x^2-1 is Minimum(-1,-3/2),Maximum(0,-1),Minimum(1,-3/2)