How do you find the domain and range of f(x) = sqrt(4+x) / (1-x)?

Aug 19, 2017

Domain: $\text{ } \left[- 4 , 1\right) \cup \left(1 , + \infty\right)$

Range: $\text{ } \left(- \infty , + \infty\right)$

Explanation:

The domain of the function represents all the values that $x$ can take for which $f \left(x\right)$ is defined.

Right from the start, you should be able to say that the domain of the function cannot include $x = 1$ because that would make the function undefined.

$1 - x \ne 0 \implies x \ne 1$

Moreover, notice that the function contains the square root of an expression that depends on the value of $x$. As you know, when working with real numbers, you cannot take the square root of a negative number.

This implies that you need

$4 + x \ge 0 \implies x \ge - 4$

Therefore, you can say that the domain of the function will be

$x \in \left[- 4 , 1\right) \cup \left(1 , + \infty\right)$

THis tells you that the function is defined for any value of $x$ that it greater than or equal to $- 4$ and is not equal to $1$.

The range of the function tells you all the possible values that $f \left(x\right)$ can take for valid domain values.

In this case, the square root of a positive number will produce a positive number, which means that regardless what value of $x$ you plug in from the domain of the function, you will have

$\sqrt{4 - x} \ge 0$

Now, for any value of $x \in \left[- 4 , 1\right)$, you will have

$\left\{\begin{matrix}\sqrt{4 + x} \ge 0 \\ 1 - x > 0\end{matrix}\right. \implies f \left(x\right) \ge 0$

and for any value of $x \in \left(1 , + \infty\right)$, you will have

$\left\{\begin{matrix}\sqrt{4 + x} > 0 \\ 1 - x < 0\end{matrix}\right. \implies f \left(x\right) < 0$

This means that the range of the function is

$\left(- \infty , 0\right] \cup \left(0 , + \infty\right) = \left(- \infty , + \infty\right)$

graph{sqrt(4+x)/(1-x) [-10, 10, -5, 5]}