# What is the domain and range of f(x) = 2 - e ^ (x / 2)?

Feb 21, 2018

Domain: $\left(- \infty , \infty\right)$

Range: $\left(- \infty , 2\right)$

#### Explanation:

The domain is all possible values of $x$ with which $f \left(x\right)$ is defined.

Here, any value of $x$ will result in a defined function. Therefore, the domain is $- \infty <$$x <$$\infty$, or, in interval notation:

$\left(- \infty , \infty\right)$.

The range is all possible values of $f \left(x\right)$. It can also be defined as the domain of ${f}^{-} 1 \left(x\right)$.

So to find ${f}^{-} 1 \left(x\right) :$

$y = 2 - {e}^{\frac{x}{2}}$

Interchange the variables $x$ and $y$:

$x = 2 - {e}^{\frac{y}{2}}$

And solve for $y$:

$x - 2 = - {e}^{\frac{y}{2}}$

${e}^{\frac{y}{2}} = 2 - x$

Take the natural logarithm of both sides:

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

$\frac{y}{2} \ln \left(e\right) = \ln \left(2 - x\right)$

As $\ln \left(e\right) = 1$,

$\frac{y}{2} = \ln \left(2 - x\right)$

$y = 2 \ln \left(2 - x\right) = {f}^{-} 1 \left(x\right)$

We must find the domain of the above.

For any $\ln x ,$ $x > 0$.

So here, $2 - x > 0$

$- x > - 2$

$x$$<$$2$

So the range of $f \left(x\right)$ can be stated as $\left(- \infty , 2\right)$