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

Oct 22, 2017

Domain $x \in \left(- \infty , + \infty\right)$ and Range $f \left(x\right) \in \left(0 , + \infty\right)$

#### Explanation:

We have the function $f \left(x\right) = {2}^{- x}$

The domain is the set of all possible $x$-values which will make the function "work", and will output real $f \left(x\right)$-values.

So in the given function any real number value of $x$ will output real $f \left(x\right)$ values.
Therefore the domain is $\mathbb{R}$ (all real values).
In interval notation it can be written as $x \in \left(- \infty , + \infty\right)$

The range is the resulting $f \left(x\right)$-values we get after substituting all the possible $x$-values
We know $x$ can be any real number.

$\textcolor{red}{1.}$So let $x =$ a negative number.

Then the function becomes $f \left(x\right) = {2}^{- \left(- x\right)}$

$f \left(x\right) = {2}^{x}$ which will always be a positive number.

$\textcolor{red}{2.}$Now let $x =$ a positive number.

Then the function becomes $f \left(x\right) = {2}^{- x}$ which can be written as $f \left(x\right) = \frac{1}{2} ^ x$.

Now as $x$ increases $f \left(x\right) \to 0$ but $f \left(x\right)$ will never be $0$.

From this we can conclude that $f \left(x\right)$ can only be a positive number inbetween $0$ and $+ \infty$ but cannot be $0$.

Therefore $f \left(x\right) \in \left(0 , \infty\right)$ and we use this ( ) bracket because neither $0$ or $\infty$ are included in the values of $f \left(x\right)$ but all the values in between $0$ and $\infty$ are included in the values of $f \left(x\right)$ aka the Range.