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

The extreme y=x^4-3x^2=x^2(x^2-3) is Minimum(-sqrt(3/2),-9/4),Maximum(0,0),Minimum(sqrt(3/2),-9/4)

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extreme y=x^4-3x^2=x^2(x^2-3)

extreme points

y = x^{4} - 3 x^{2} = x^{2} (x^{2} - 3)

Minimum (- \frac{3}{2}, - \frac{9}{4}), Maximum (0, 0), Minimum (\frac{3}{2}, - \frac{9}{4})

Solution steps

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

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^{2} (x^{2} - 3) : - \frac{9}{4}

Minimum (- \frac{3}{2}, - \frac{9}{4})

Plug

x = 0

into

x^{2} (x^{2} - 3) : 0

Maximum (0, 0)

Plug

x = \frac{3}{2}

into

x^{2} (x^{2} - 3) : - \frac{9}{4}

Minimum (\frac{3}{2}, - \frac{9}{4})

Minimum (- \frac{3}{2}, - \frac{9}{4}), Maximum (0, 0), Minimum (\frac{3}{2}, - \frac{9}{4})

f (x) = x^{2} + y^{2} + x^{2} y + 4

f (x) = \frac{1}{3} x^{3} + 5 x^{2} + 24 x + 7

f (x) = \frac{x ^{2} - 16 x + 64}{x - 10}

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

f (x) = x^{3} + y^{3} - 33 x y

What is the extreme y=x^4-3x^2=x^2(x^2-3) ?
The extreme y=x^4-3x^2=x^2(x^2-3) is Minimum(-sqrt(3/2),-9/4),Maximum(0,0),Minimum(sqrt(3/2),-9/4)