# Restrictions on Domain

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At this stage, we will only discuss functions for which the set of input values is   $$\mathbb{R}$$ or a subset of $$\mathbb{R}$$ . That is, we won’t talk about functions in which the input variable is complex-valued.

Consider the function $$f\left( x \right) = \frac{1}{x}$$. For any input value, this function outputs the reciprocal of the input value. Think about this: among the set of all Real numbers, what values can x take?

The answer is simple: as long as is any non-zero real number, its reciprocal is well-defined. However, if is 0, then $$\frac{1}{x}$$ is a mathematically undefined / invalid entity. Thus, x can take on any real value other than 0. This means that the domain of $$f\left( x \right)$$ is the set of all real numbers except 0:

$D=\mathbb{R}-\left\{0\right\}$

Note that a restriction had to be placed on the possible set of input values due to the nature of the function.These kinds of restrictions can arise in many situations, where the domain could be a subset of   $$\mathbb{R}$$  rather than the entire set $$\mathbb{R}$$.

As another example, consider the function.

$f\left( x \right) = \sqrt {x - 1}$

Suppose that we want the output of to be a real number. What restriction does this requirement place on the set of input values?

The output will be real only if the term under the square root sign is non-negative. Thus, we require:

$x - 1 \ge 0\,\,\, \Rightarrow \,\,\,x \ge 1$

Using interval notation, we can write the domain of f as $$D = \left[ {1,\infty} \right)$$.

You must understand that this restriction (on the possible values which x can take) arises in this case because we have restricted the output of f to be real-valued. If that requirement is lifted, x can take any other real value outside $$\left[ {1,\infty } \right)$$also – the output will be non-real in that case (in fact, x itself can take non-real values, but as we said earlier, we are not considering such scenarios at this stage).

To summarize, a restriction on the domain can arise because of the following reasons:

1. Then nature of the function requires the input values to be restricted. For example,the input variable should not take a value such that the denominator of a fractional term becomes 0.

2. The output set has been restricted. Sometimes, this restriction may be specified explicitly in the question. Otherwise, at your level, we will always follow the restriction that the output values must lie in the Real set – that is, we will only discuss real-      valued functions, or functions which generate real-valued outputs. This is more because we want to keep things simple at this stage, rather than any other reason. Thus, whenever we talk about a function, the assumption that it is a real-valued      function   is  implicit.

Example 1: Suppose that we want the output of the function $$f\left( x \right) = \frac{x}{2}$$ to be an integer. What can be the largest possible domain of f?

Solution: For $$\frac{x}{2}$$ to be an integer, x must be an even integer. Thus, the largest possible domain of f is the set of all even integers.

Note. Whenever we say something like“Find the domain of f”, it should be interpreted as “Find the largest possible set of real input values for f so that f generates real-valued outputs”.

Example 2: Find the domain of.

$f\left( x \right) = \frac{x}{{{x^2} - 3x + 2}}$

Solution: The denominator in the expression must not be 0, otherwise the output value will be mathematically undefined. Thus,

$\begin{array}{l}{x^2} - 3x + 2 \ne 0\\ \Rightarrow \,\,\,\left( {x - 1}\right)\left( {x - 2} \right) \ne 0\\ \Rightarrow \,\,\,x \ne 1,2\end{array}$

This means that the domain of f  is $$\mathbb{R}- \left\{ {1,2} \right\}$$ .

Example 3: Find the domain of

$g\left( y \right) = \sqrt {{y^2} - 4y + 3}$

Solution: The input variable y must only take on such values for which the expression under the square root sign is non-negative, otherwise the function g will generate non-real values.Thus, we have:

$\begin{array}{l}{y^2} - 4y + 3 \ge 0\\ \Rightarrow \,\,\,\left( {y - 1}\right)\left( {y - 3} \right) \ge 0\\ \Rightarrow \,\,\,y \le1\,\,\,{\rm{or}}\,\,\,y \ge 3\end{array}$

We can write the domain of the function as

$D = \left( { - \infty ,1} \right] \cup \left[{3,\infty } \right)$

Example 4: Find the domain of.

$f\left( x \right) = \frac{1}{{{x^2} - 1}} +\frac{2}{{4 - {x^2}}}$

Solution: The following restriction needsto be applied:

$\begin{array}{l}{x^2} \ne 1,4\\ \Rightarrow \,\,\,x \ne - 1,1, -2,2\end{array}$

Thus, the domain of the function can be written as:

$D = \mathbb{R} - \left\{ { - 2, - 1,1,2} \right\}$

Example 5: Find the domain of.

$f\left( x \right) = {x^2} - 5x + 4$

so that f generates only non-negative real outputs.

Solution: A restriction on the set of outputs has been placed artificially in the problem. We have:

$\begin{array}{l}{x^2} - 5x + 4 \ge 0\\ \Rightarrow \,\,\,\left( {x - 1}\right)\left( {x - 4} \right) \ge 0\\ \Rightarrow \,\,\,x \le1\,\,\,{\rm{or}}\,\,\,x \ge 4\end{array}$

Thus, the domain of this function is

$D = \left( { - \infty ,1} \right] \cup \left[{4,\infty } \right)$

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