What is the extreme f(x)=(x-2)/(x^2+3x+6) ?

The extreme f(x)=(x-2)/(x^2+3x+6) is Minimum(-2,-1),Maximum(6, 1/15)

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extreme f(x)=(x-2)/(x^2+3x+6)

extreme points

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

Minimum (- 2, - 1), Maximum (6, \frac{1}{15})

Solution steps

f^{'} (x) = \frac{- x ^{2} + 4 x + 12}{( x ^{2} + 3 x + 6 ) ^{2}}

Find intervals:

Decreasing

: - \infty < x < - 2,

Increasing

: - 2 < x < 6,

Decreasing

: 6 < x < \infty

Plug

x = - 2

into

\frac{x - 2}{x ^{2} + 3 x + 6} : - 1

Minimum (- 2, - 1)

Plug

x = 6

into

\frac{x - 2}{x ^{2} + 3 x + 6} : \frac{1}{15}

Maximum (6, \frac{1}{15})

Minimum (- 2, - 1), Maximum (6, \frac{1}{15})

\frac{6 - x}{x - 4}

y = - x - 7

(- 3, 3), (4, 4)

- 8 csc (\frac{π}{3} x)

f (x) = \frac{- 6}{x + 3}

What is the extreme f(x)=(x-2)/(x^2+3x+6) ?
The extreme f(x)=(x-2)/(x^2+3x+6) is Minimum(-2,-1),Maximum(6, 1/15)