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

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

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

extreme points

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

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

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < - 3,

Increasing

: - 3 < x < 0,

Decreasing

: 0 < x < 3,

Increasing

: 3 < x < \infty

Plug

x = - 3

into

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

Minimum (- 3, 0)

Plug

x = 0

into

(x^{2} - 3)^{2} : 9

Maximum (0, 9)

Plug

x = 3

into

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

Minimum (3, 0)

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

f (x) = (x - 1) e^{x}

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

f (x, y) = - x^{2} - y^{2} + 8 x + 6 y

f (x) = x + e^{- 2 x}

f (x) = \frac{x + 1}{x ^{2} + x + 1}

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