Express the function $ f $ without using absolute value signs $\left|\frac{x-2}{x+3}\right|e^{\left|x-2\right|}$?

Question

Good evening to everyone: This is the equation $$ f(x) = \left|\frac{x-2}{x+3}\right|e^{\left|x-2\right|} $$ What I've tried is: $$ \frac{x-2}{x+3}\ge 0 => x-2 \ge 0 => x \ge 2$$ Then $$ \frac{-x+2}{-x-3} < 0 => -x+2 <0 => x > 2$$ These 2 combined give the result $$ \left|\frac{x-2}{x+3}\right| = \frac{-x+2}{-x-3} $$ for $x>2 $ and $$x-2 \ge 0 => x \ge 2$$ and $$ -x+2<0 => -x<-2 => x>2$$ therefore the result is $$ \left|x-2\right| =-x+2$$ for $x>2$. So the equation becomes $$ f(x) = \frac{-x+2}{-x-3}e^{-x+2} $$ for $ x>2 $ .But my teacher says it's not right. Can you clarify it for me please. Thanks for any possible response. Edit: The real answers are: $$\frac{x-2}{x+3}e^{2-x}\:$$ for $ x<3$ $$ \frac{2-x}{x+3}e^{2-x}\: $$ for $-3 < x \le 2 $ and $$ \frac{x-2}{x+3}e^{x-2} $$ for $ x > 2 $

Answer 1

$$\frac{x-2}{x+3}\ge 0\stackrel{\text{Mult. by}\;(x+3)^2}\iff (x-2)(x+3)\ge0\;,\;\;x\neq-3\iff$$

$$x<-3\;\;\text{or}\;\;x\ge2$$

and then

$$\left|\frac{x-2}{x+3}\right|e^{|x-2|}=\begin{cases}\frac{x-2}{x+3}e^{x-2},&x\ge2\\{}\\\frac{x-2}{x+3}e^{-(x-2)},&x<-3\end{cases}$$

And thus

$$-3<x<2\implies \left|\frac{x-2}{x+3}\right|e^{|x-2|}=-\frac{x-2}{x+3}e^{-(x-2)}$$

Answer 2

Hint:

1) $$\left|\frac{x-2}{x+3} \right|=\frac{x-2}{x+3} \Leftrightarrow x\in (-\infty;-3) \cup [2;+\infty)$$

2)$$\left|\frac{x-2}{x+3} \right|=-\frac{x-2}{x+3}\Leftrightarrow x\in (-3;2)$$