1. ## Critical value

Hi.
There is no critical value for 1/(x+1)^2.Right?
The open interval is (-∞,∞).

2. Originally Posted by wowo
Hi.
There is no critical value for 1/(x+1)^2.Right?
The open interval is (-∞,∞).
Are you talking about doing a derivative test?

3. Originally Posted by wowo
Hi.
There is no critical value for 1/(x+1)^2.Right?
The open interval is (-∞,∞).
Also, the interval is $\displaystyle (-\infty,-1)\cup (-1,\infty)$

4. Yes

5. Originally Posted by wowo
Yes
$\displaystyle \displaystyle\frac{d}{dx}\left[\frac{1}{(x+1)^2}\right]=\frac{-2}{(x+1)^3}$

$\displaystyle \displaystyle \frac{-2}{(x+1)^3}=0$

Critical value is -1

6. f(x)=x/(x+1)^2
f'(x)=1/(x+1)^2

Sorry.

7. Originally Posted by wowo
f(x)=x/(x+1)^2
f'(x)=1/(x+1)^2

Sorry.
$\displaystyle \displaystyle f'(x)=\frac{-(x-1)}{(x+1)^2}=0\Rightarrow x-1=0 \ \mbox{and} \ x\neq -1$

8. Originally Posted by dwsmith
$\displaystyle \displaystyle f'(x)=\frac{-(x-1)}{(x+1)^2}=0\Rightarrow x-1=0 \ \mbox{and} \ x\neq -1$
The interval is (-∞,∞)?

9. Originally Posted by wowo
The interval is (-∞,∞)?
No, $\displaystyle (-\infty, -1)\cup (-1,\infty)$

If x = -1, you will be dividing by 0.