What is the extreme f(x)=x^4-7x^2+8 ?

The extreme f(x)=x^4-7x^2+8 is Minimum(-sqrt(7/2),-17/4),Maximum(0,8),Minimum(sqrt(7/2),-17/4)

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extreme f(x)=x^4-7x^2+8

extreme points

f (x) = x^{4} - 7 x^{2} + 8

Minimum (- \frac{7}{2}, - \frac{17}{4}), Maximum (0, 8), Minimum (\frac{7}{2}, - \frac{17}{4})

Solution steps

f^{'} (x) = 4 x^{3} - 14 x

Find intervals:

Decreasing

: - \infty < x < - \frac{7}{2},

Increasing

: - \frac{7}{2} < x < 0,

Decreasing

: 0 < x < \frac{7}{2},

Increasing

: \frac{7}{2} < x < \infty

Plug

x = - \frac{7}{2}

into

x^{4} - 7 x^{2} + 8 : - \frac{17}{4}

Minimum (- \frac{7}{2}, - \frac{17}{4})

Plug

x = 0

into

x^{4} - 7 x^{2} + 8 : 8

Maximum (0, 8)

Plug

x = \frac{7}{2}

into

x^{4} - 7 x^{2} + 8 : - \frac{17}{4}

Minimum (\frac{7}{2}, - \frac{17}{4})

Minimum (- \frac{7}{2}, - \frac{17}{4}), Maximum (0, 8), Minimum (\frac{7}{2}, - \frac{17}{4})

What is the extreme f(x)=x^4-7x^2+8 ?
The extreme f(x)=x^4-7x^2+8 is Minimum(-sqrt(7/2),-17/4),Maximum(0,8),Minimum(sqrt(7/2),-17/4)