Critical value

**wowo** · January 17th 2011, 06:19 PM

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

**dwsmith** · January 17th 2011, 06:19 PM

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?

**dwsmith** · January 17th 2011, 06:21 PM

Originally Posted by wowo

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

Also, the interval is $(-\infty,-1)\cup (-1,\infty)$

**wowo** · January 17th 2011, 06:22 PM

Yes

**dwsmith** · January 17th 2011, 06:28 PM

Originally Posted by wowo

Yes

$\displaystyle\frac{d}{dx}\left[\frac{1}{(x+1)^2}\right]=\frac{-2}{(x+1)^3}$

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

Critical value is -1

**wowo** · January 17th 2011, 06:31 PM

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

My bad.
Sorry.

**dwsmith** · January 17th 2011, 06:33 PM

Originally Posted by wowo

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

My bad.
Sorry.

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

**wowo** · January 17th 2011, 06:40 PM

Originally Posted by dwsmith

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

The interval is (-∞,∞)?

**dwsmith** · January 17th 2011, 06:42 PM

Originally Posted by wowo

The interval is (-∞,∞)?

No, $(-\infty, -1)\cup (-1,\infty)$

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

Thread: Critical value