How to solve $xe^{\frac{x-1}{x+1}}>0$?

Question

How to solve $xe^{\frac{x-1}{x+1}}>0$?

Please give me a hint, I don't know where to start.

Answer 1

Note that $\mathrm{e}^{z} > 0$ for all $z \in \mathbb{R}$. Thus $$ x\mathrm{e}^{\frac{x-1}{x+1}} > 0 \iff x > 0. $$

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It is worth observing that $\mathrm{e}^{\frac{x-1}{x+1}}$ is undefined if $x = -1$, but this case causes us no problems here, as the original inequality only holds if $x > 0$. However, if we we were interested in the inequality $$ x \mathrm{e}^{\frac{x-1}{x+1}} < 0, $$ this technicality would matter. For this inequality to hold, we not that $\mathrm{e}^{z} > 0$ for all $z \in \mathbb{R}$, hence $$ \mathrm{e}^{\frac{x-1}{x+1}} > 0 \qquad \forall x\in \mathbb{R}\setminus\{-1\}. $$ Thus $$ x \mathrm{e}^{\frac{x-1}{x+1}} < 0 \iff (x < 0) \land (x \ne -1), $$ i.e. if $x \in (-\infty,-1) \cup (-1,0)$.

Answer 2

You have $x\cdot e^{\frac{x-1}{x+1}}\gt 0$

note that $e^a\gt0 \quad\forall a\in \mathbf R $

Hence your inequality reduces to $x \gt 0 $

Answer 3

Note that:

$e^{\frac{x-1}{x+1}}>0 \quad \forall x \ne -1$

so the product $xe^{\frac{x-1}{x+1}}$ is positive if $x$ is positive

Answer 4

A product is positive if and only if either both factors are positive or both factors are negative. Can the latter situation happen?

Don't forget that the factors need to exist and check for it.

Answer 5

Since exp((x-1)/(x+1))is always positive if it does exist,then the solution set of this inequality shall be x>0, as you can see,exp((x-1)/(x+1)) always exists if x doesn't equal -1.