# Sign of the derivative $-e^{\frac{1}{2x+2}}\left(\operatorname{sgn}\left(x\right)+\frac{1-\left|x\right|}{2\left(x+1\right)^2}\right)$

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!

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 $-1<x<-1/2$.