What is the extreme f(x)=x^4-6x^2+9 ?

The extreme f(x)=x^4-6x^2+9 is Minimum(-sqrt(3),0),Maximum(0,9),Minimum(sqrt(3),0)

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extreme f(x)=x^4-6x^2+9

extreme points

f (x) = x^{4} - 6 x^{2} + 9

Minimum (- 3, 0), Maximum (0, 9), Minimum (3, 0)

Solution steps

f^{'} (x) = 4 x^{3} - 12 x

Find intervals:

Decreasing

: - \infty < x < - 3,

Increasing

: - 3 < x < 0,

Decreasing

: 0 < x < 3,

Increasing

: 3 < x < \infty

Plug

x = - 3

into

x^{4} - 6 x^{2} + 9 : 0

Minimum (- 3, 0)

Plug

x = 0

into

x^{4} - 6 x^{2} + 9 : 9

Maximum (0, 9)

Plug

x = 3

into

x^{4} - 6 x^{2} + 9 : 0

Minimum (3, 0)

Minimum (- 3, 0), Maximum (0, 9), Minimum (3, 0)

f (x) = - x^{2} + 8 x - 15

f (x) = x^{\frac{4}{5}} (9 - 4 x), 0 \leq x \leq 2

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

f (x) = 2 x^{3} - 13 x^{2} + 24 x - 28

f (x, y) = x^{3} - y^{3} - 2 x y + 7

What is the extreme f(x)=x^4-6x^2+9 ?
The extreme f(x)=x^4-6x^2+9 is Minimum(-sqrt(3),0),Maximum(0,9),Minimum(sqrt(3),0)