How to Find Values of x Outside the Domain of h

To list the values of $x$ that are not in the domain of $h$, you need the expression that defines $h$. Without it, there is no unique set of excluded values: a polynomial has no real exclusions, while a rational function or logarithm may exclude some inputs.

The method is to find every input that makes any part of the formula undefined, then combine those restrictions. The examples below show how to do that for common real-valued functions.

What “outside the domain” means

The domain is the set of allowed inputs. Values outside the domain are its complement: every candidate input that the function’s definition does not permit. Unless a problem specifies otherwise, algebra and precalculus problems usually take the inputs to be real numbers.

For example, if the domain is $[0,\infty)$, then the real values outside it are $(-\infty,0)$. The exact answer depends on the function’s formula and on the assumed input set.

Check each operation in h

Look for the following restrictions in the original expression:

Part of the expression	Condition for real-valued inputs
A fraction with denominator $q(x)$	$q(x) \ne 0$
An even root, such as $\sqrt{r(x)}$	$r(x) \ge 0$
A logarithm, such as $\ln(r(x))$	$r(x) > 0$
A composition, such as $f(g(x))$	$g(x)$ must be defined and must be in the domain of $f$

Odd roots, such as $\sqrt[3]{x}$, allow negative, zero, and positive real inputs. A polynomial also has no real-domain restriction. Other function types can have their own restrictions, so check all parts of the given formula.

Worked example

Suppose the intended function is

$$h(x)=\frac{\sqrt{2x-1}}{x-3}+\ln(x+2).$$

Check each potentially restricted part separately:

1. The square root requires $2x-1\ge 0$, so $x\ge \frac{1}{2}$.
2. The denominator requires $x-3\ne 0$, so $x\ne 3$.
3. The logarithm requires $x+2>0$, so $x>-2$.

All three conditions must hold at once. The strongest lower bound is $x\ge \frac{1}{2}$, and $x=3$ must be left out. Thus

$$\operatorname{Dom}(h)=\left[\frac{1}{2},\infty\right)\setminus\{3\}.$$

The real numbers not in this domain are

$$(-\infty,\tfrac{1}{2})\cup\{3\}.$$

This illustrates an important point: the excluded values are the union of all values ruled out by the individual restrictions. The logarithm by itself excludes $x\le -2$, but those inputs were already excluded by the square-root condition in this example.

Common mistakes

• Checking only the denominator. A formula can also contain a square root, logarithm, or other restricted operation.
• Using the wrong inequality. For an even root, zero is allowed; for a logarithm, an argument of zero is not.
• Forgetting that every condition must hold. Combine restrictions by intersection to find the domain, or take their union to list excluded inputs.
• Simplifying away a restriction. For example, $\frac{x^2-1}{x-1}$ is undefined at $x=1$ in its original form. Although it simplifies to $x+1$ when $x\ne1$, the original function still excludes $1$.
• Assuming the input set. If a problem uses complex inputs or specifies a restricted set, the answer may differ from the usual real-domain answer.

To find the exact values for a particular $h$, write down its definition and apply these checks to every part. Without that definition, only the general method—not a specific list of excluded $x$-values—can be determined.