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

The extreme (x^2+x-2)/(x+3) is Maximum(-5,-9),Minimum(-1,-1)

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

extreme points

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

Maximum (- 5, - 9), Minimum (- 1, - 1)

Solution steps

f^{'} (x) = \frac{x ^{2} + 6 x + 5}{( x + 3 ) ^{2}}

Find intervals:

Increasing

: - \infty < x < - 5,

Decreasing

: - 5 < x < - 3,

Decreasing

: - 3 < x < - 1,

Increasing

: - 1 < x < \infty

Plug

x = - 5

into

\frac{x ^{2} + x - 2}{x + 3} : - 9

Maximum (- 5, - 9)

Plug

x = - 1

into

\frac{x ^{2} + x - 2}{x + 3} : - 1

Minimum (- 1, - 1)

Maximum (- 5, - 9), Minimum (- 1, - 1)

f (x, y) = e^{x} ln (1 + y)

y = x^{\frac{2}{7}} (x^{2} - 4)

x^{2} e^{x^{2}}

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

f (x) = x^{2} - 4 x - 6 [0.2]

What is the extreme (x^2+x-2)/(x+3) ?
The extreme (x^2+x-2)/(x+3) is Maximum(-5,-9),Minimum(-1,-1)