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

The extreme f(x)=(x^2-3)/(x+2) is Maximum(-3,-6),Minimum(-1,-2)

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

extreme points

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

Maximum (- 3, - 6), Minimum (- 1, - 2)

Solution steps

f^{'} (x) = \frac{x ^{2} + 4 x + 3}{( x + 2 ) ^{2}}

Find intervals:

Increasing

: - \infty < x < - 3,

Decreasing

: - 3 < x < - 2,

Decreasing

: - 2 < x < - 1,

Increasing

: - 1 < x < \infty

Plug

x = - 3

into

\frac{x ^{2} - 3}{x + 2} : - 6

Maximum (- 3, - 6)

Plug

x = - 1

into

\frac{x ^{2} - 3}{x + 2} : - 2

Minimum (- 1, - 2)

Maximum (- 3, - 6), Minimum (- 1, - 2)

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

f (x) = - 0.12 x^{2} + 168 x + 200

f (x) = \frac{ln ( x )}{x ^{4}}

\frac{2 x ^{2}}{x ^{2} - 9}

x^{3} ln (x)

What is the extreme f(x)=(x^2-3)/(x+2) ?
The extreme f(x)=(x^2-3)/(x+2) is Maximum(-3,-6),Minimum(-1,-2)