# How do you find the domain and range of f(x) = 1/(2-e^x)?

Aug 3, 2015

Domain: $\left(- \infty , \ln \left(2\right)\right) \cup \left(\ln \left(2\right) , + \infty\right)$
Range: $\left(- \infty , 0\right) \cup \left(\frac{1}{2} , + \infty\right)$

#### Explanation:

Right from the start, you can say that one value of $x$ will be excluded from the domain of the function because of the restriction to the denominator of the function.

More epecifically, the expression $\left(2 - {e}^{x}\right)$ must not be equal to zero.

The value of $x$ for which condition is not satisfied will be

$2 - {e}^{x} = 0$

${e}^{x} = 2$

$\ln \left({e}^{x}\right) = \ln \left(2\right)$

$x \cdot \ln \left(e\right) = \ln \left(2\right) \implies x = \ln \left(2\right)$

The domain of the function will thus be $\mathbb{R} - \left\{\ln \left(2\right)\right\}$, or $\left(- \infty , \ln \left(2\right)\right) \cup \left(\ln \left(2\right) , + \infty\right)$.

The range of the function will be affected by the fact that the graph has a vertical asymptote at $x = \ln \left(2\right)$.

Now, because ${e}^{x} > 0 , \left(\forall\right) x$, you get that

$\frac{1}{2 - {e}^{x}} > 0$ for $x < \ln \left(2\right)$ and $\frac{1}{2 - {e}^{x}} < 0$ for $x > \ln \left(2\right)$.

That happens because you have

${\lim}_{x \to - \infty} {e}^{x} = 0$, for which you have $f \left(x\right) \to \frac{1}{2}$

${\lim}_{x \to \infty} {e}^{x} = + \infty$, for which you have $f \left(x\right) \to 0$

The range of your function will thus be $\left(- \infty , 0\right) \cup \left(\frac{1}{2} , + \infty\right)$

graph{1/(2-e^x) [-10, 10, -5, 5]}