Hi. There is no critical value for 1/(x+1)^2.Right? The open interval is (-∞,∞).
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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?
Originally Posted by wowo Hi. There is no critical value for 1/(x+1)^2.Right? The open interval is (-∞,∞). Also, the interval is $\displaystyle (-\infty,-1)\cup (-1,\infty)$
Yes
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
f(x)=x/(x+1)^2 f'(x)=1/(x+1)^2 My bad. Sorry.
Originally Posted by wowo f(x)=x/(x+1)^2 f'(x)=1/(x+1)^2 My bad. Sorry. $\displaystyle \displaystyle f'(x)=\frac{-(x-1)}{(x+1)^2}=0\Rightarrow x-1=0 \ \mbox{and} \ x\neq -1$
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 (-∞,∞)?
Originally Posted by wowo The interval is (-∞,∞)? No, $\displaystyle (-\infty, -1)\cup (-1,\infty)$ If x = -1, you will be dividing by 0.
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