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)

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

\frac{1}{3} (x - 2.1)^{2} + 7.9

f (x, y) = 120 x + 120 y - x y - x^{2} - y^{2}

f (x) = (x + 7)^{\frac{2}{3}}

f (x) = - x + 2 cos (x)

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)