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

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)

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)