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

The extreme f(x)=x^4-x^2+2 is Minimum(-1/(sqrt(2)), 7/4),Maximum(0,2),Minimum(1/(sqrt(2)), 7/4)

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

extreme

Minimum (- \frac{1}{2}, \frac{7}{4}), Maximum (0, 2), Minimum (\frac{1}{2}, \frac{7}{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} + 2 : \frac{7}{4}

Minimum (- \frac{1}{2}, \frac{7}{4})

Plug

x = 0

into

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

Maximum (0, 2)

Plug

x = \frac{1}{2}

into

x^{4} - x^{2} + 2 : \frac{7}{4}

Minimum (\frac{1}{2}, \frac{7}{4})

Minimum (- \frac{1}{2}, \frac{7}{4}), Maximum (0, 2), Minimum (\frac{1}{2}, \frac{7}{4})

What is the extreme f(x)=x^4-x^2+2 ?
The extreme f(x)=x^4-x^2+2 is Minimum(-1/(sqrt(2)), 7/4),Maximum(0,2),Minimum(1/(sqrt(2)), 7/4)