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

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)

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)