What is the extreme f(x)=((x^2-8))/(x+3) ?

The extreme f(x)=((x^2-8))/(x+3) is Maximum(-4,-8),Minimum(-2,-4)

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extreme f(x)=((x^2-8))/(x+3)

extreme points

f (x) = \frac{( x ^{2} - 8 )}{x + 3}

Maximum (- 4, - 8), Minimum (- 2, - 4)

Solution steps

f^{'} (x) = \frac{x ^{2} + 6 x + 8}{( x + 3 ) ^{2}}

Find intervals:

Increasing

: - \infty < x < - 4,

Decreasing

: - 4 < x < - 3,

Decreasing

: - 3 < x < - 2,

Increasing

: - 2 < x < \infty

Plug

x = - 4

into

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

Maximum (- 4, - 8)

Plug

x = - 2

into

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

Minimum (- 2, - 4)

Maximum (- 4, - 8), Minimum (- 2, - 4)

y = \frac{3}{- ( x - 4 ) ^{2} - 1}

f (x y) = 2 x^{2} + 2 y^{2} + 4 x y - 8 x + 20

f (x) = ln (3 - ln (x))

f (x) = 12 x^{2} - 2 x^{3}

f (x, y) = x^{2} + y^{2} - 4 y + 4

What is the extreme f(x)=((x^2-8))/(x+3) ?
The extreme f(x)=((x^2-8))/(x+3) is Maximum(-4,-8),Minimum(-2,-4)