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, 1/2),Maximum(0,1),Minimum(1, 1/2)

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

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

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

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

(- 3.5, 15), (6, 13.5)

f (x) = e^{- x}

f (x) = 9 - 4 x

f (x) = lo g_{5} (x + 3)

f (x) = \frac{- 1}{x}

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, 1/2),Maximum(0,1),Minimum(1, 1/2)