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

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

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

extreme points

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

Minimum (- 5, - \frac{2}{9}), Maximum (- 1, - 2)

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < - 5,

Increasing

: - 5 < x < - 2,

Increasing

: - 2 < x < - 1,

Decreasing

: - 1 < x < 1,

Decreasing

: 1 < x < \infty

Plug

x = - 5

into

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

Minimum (- 5, - \frac{2}{9})

Plug

x = - 1

into

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

Maximum (- 1, - 2)

Minimum (- 5, - \frac{2}{9}), Maximum (- 1, - 2)

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

f (x) = 2 x^{3} - 33 x^{2} + 180 x + 6

f (x) = (x^{4} - 32 x^{2} - 4)

f (x) = 5 x^{3} e^{- x}, - 1 \leq x \leq 4

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

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