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

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

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

extreme points

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

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

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

Plug

x = 0

into

x^{4} - 3 x^{2} : 0

Maximum (0, 0)

Plug

x = \frac{3}{2}

into

x^{4} - 3 x^{2} : - \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) = \frac{2 x + 8}{x ^{2} + 3 x - 4}

lo g_{10} (2 x + 5)

f (x) = \frac{9 x + 4}{x - 7}

f (x) = 3 x^{2} - 5 x - 4

f (x) = \frac{3}{2 x - 5}

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