What is the extreme f(x)=x^4-3x^2+2 ?

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

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extreme f(x)=x^4-3x^2+2

extreme points

f (x) = x^{4} - 3 x^{2} + 2

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

Solution steps

f^{'} (x) = 4 x^{3} - 6 x

Find intervals:

Decreasing

: - \infty < x < - \frac{3}{2},

Increasing

: - \frac{3}{2} < x < 0,

Decreasing

: 0 < x < \frac{3}{2},

Increasing

: \frac{3}{2} < x < \infty

Plug

x = - \frac{3}{2}

into

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

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

Plug

x = 0

into

x^{4} - 3 x^{2} + 2 : 2

Maximum (0, 2)

Plug

x = \frac{3}{2}

into

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

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

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

in v erse f (x) = \frac{4 x}{7 x - 1}

d i s t an ce (- 3, 7), (- 8, 5)

s im pl i f y (9.1) (3.4)

s l o p e in t erce pt x - 4 y = - 1

p er i o d i c i t y f (x) = cos (\frac{n}{4})

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