What is the extreme f(x)=(x-1)/(x^2+5x+10) ?

The extreme f(x)=(x-1)/(x^2+5x+10) is Minimum(-3,-1),Maximum(5, 1/15)

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extreme f(x)=(x-1)/(x^2+5x+10)

extreme points

f (x) = \frac{x - 1}{x ^{2} + 5 x + 10}

Minimum (- 3, - 1), Maximum (5, \frac{1}{15})

Solution steps

f^{'} (x) = \frac{- x ^{2} + 2 x + 15}{( x ^{2} + 5 x + 10 ) ^{2}}

Find intervals:

Decreasing

: - \infty < x < - 3,

Increasing

: - 3 < x < 5,

Decreasing

: 5 < x < \infty

Plug

x = - 3

into

\frac{x - 1}{x ^{2} + 5 x + 10} : - 1

Minimum (- 3, - 1)

Plug

x = 5

into

\frac{x - 1}{x ^{2} + 5 x + 10} : \frac{1}{15}

Maximum (5, \frac{1}{15})

Minimum (- 3, - 1), Maximum (5, \frac{1}{15})

What is the extreme f(x)=(x-1)/(x^2+5x+10) ?
The extreme f(x)=(x-1)/(x^2+5x+10) is Minimum(-3,-1),Maximum(5, 1/15)