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

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

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

extreme points

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

Minimum (- 2, 0), Minimum (2, 0)

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < - 2,

Increasing

: 2 < x < \infty

Plug

x = - 2

into

x^{2} - 4 : 0

Minimum (- 2, 0)

Plug

x = 2

into

x^{2} - 4 : 0

Minimum (2, 0)

Minimum (- 2, 0), Minimum (2, 0)

\frac{d}{d x} (I n \frac{x + 1}{x - 1})

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

f (x) = \frac{x ^{3}}{x ^{2} - 9}

y = - x^{3} - 3 x + 4

12 - b + c

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