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Popular >

extreme f(x)= 1/(x^2+2x+9)

Solution

extreme points

f (x) = \frac{1}{x ^{2} + 2 x + 9}

Solution

Maximum (- 1, \frac{1}{8})

Solution steps

f^{'} (x) = - \frac{2 x + 2}{( x ^{2} + 2 x + 9 ) ^{2}}

Find intervals:

Increasing

: - \infty < x < - 1,

Decreasing

: - 1 < x < \infty

Plug

x = - 1

into

\frac{1}{x ^{2} + 2 x + 9} : \frac{1}{8}

Maximum (- 1, \frac{1}{8})

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Frequently Asked Questions (FAQ)

What is the extreme f(x)= 1/(x^2+2x+9) ?
The extreme f(x)= 1/(x^2+2x+9) is Maximum(-1, 1/8)