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

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

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

extreme points

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

Minimum (- \frac{1}{2}, \frac{3}{4}), Maximum (0, 1), Minimum (\frac{1}{2}, \frac{3}{4})

Solution steps

f^{'} (x) = 4 x^{3} - 2 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} - x^{2} + 1 : \frac{3}{4}

Minimum (- \frac{1}{2}, \frac{3}{4})

Plug

x = 0

into

x^{4} - x^{2} + 1 : 1

Maximum (0, 1)

Plug

x = \frac{1}{2}

into

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

Minimum (\frac{1}{2}, \frac{3}{4})

Minimum (- \frac{1}{2}, \frac{3}{4}), Maximum (0, 1), Minimum (\frac{1}{2}, \frac{3}{4})

f (x) = x^{3} + 6 x^{2} - 15 x + 2

x^{3} + y^{3} - 6 x y

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

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

f (x) = x - x

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