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Good morning to everyone. I have a problem with finding the sign of a derivative: $$ \frac{d}{dx}f(x)=-e^{\frac{1}{2x+2}}\left(\operatorname{sgn}\left(x\right)+\frac{1-\left|x\right|}{2\left(x+1\right)^2}\right) $$ Therefore I don't know how to find the intervals where it's increasing and where it's decreasing and the stationary points. My Solution: First I have to do its sign, therefore: $$ f(x)\ge 0 $$ For $ x < 0 $ the equation becomes $\frac{\left(1-\left|x\right|\right)e^{\frac{1}{2x+2}}}{2\left(x+1\right)^2} $ which is $ \ge 0 $ for $ 1-\left|x\right|\ge 0\: $ therefore $ x $ belongs to the interval $ [0,1] $.

For $ x > 0 $ the equation becomes $-\frac{\left(1-\left|x\right|\right)e^{\frac{1}{2x+2}}}{2\left(x+1\right)^2} $ which is $ \ge 0 $ for $ 1-\left|x\right|\le 0\: $ therefore $ x $ belongs to the interval $ [0,1] $. The function increases on the interval $ (-\infty,0] $ decreses on the interval $ [0,1] $ and increases on the interval $[1,\infty) $

But my teacher says it's not good. Why? Or at least what's the correct response? Thanks for any possible response!

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Let me assume you computed the derivative right. For $x>0$ the derivative has the same sign as $$ -\left(1+\frac{1-x}{2(x+1)^2}\right)=-\frac{2x^2+3x+3}{2(x+1)^2} $$ because the exponential factor is positive. So…

the function is decreasing for $x>0$

For $x<0$ the derivative has the same sign as $$ -\left(-1+\frac{1+x}{2(x^2+1)}\right)=\frac{2x^2+3x+1}{2(x+1)^2} $$ which is positive for…

$x<-1$ or $-1/2<x<0$, and negative for $-1<x<-1/2$.

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