What is the extreme f(x)=(x-1)/(x^2+3x) ?

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

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extreme f(x)=(x-1)/(x^2+3x)

extreme

Minimum (- 1, 1), Maximum (3, \frac{1}{9})

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < - 3,

Decreasing

: - 3 < x < - 1,

Increasing

: - 1 < x < 0,

Increasing

: 0 < x < 3,

Decreasing

: 3 < x < \infty

Plug

x = - 1

into

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

Minimum (- 1, 1)

Plug

x = 3

into

\frac{x - 1}{x ^{2} + 3 x} : \frac{1}{9}

Maximum (3, \frac{1}{9})

Minimum (- 1, 1), Maximum (3, \frac{1}{9})

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