1.5 Restrictions

When we define a function, we may specify restrictions on its input variables by stating what values are allowed or disallowed. The division component comes with a restriction: we must avoid dividing by zero.

Whenever we write a function that uses division, we'll include restrictions to ensure that the denominator is non-zero.

Example.

Consider the function

\[ \begin{align} z &\depends x \\ z &= \frac{3}{x^2 - 4} + 6. \end{align} \]

What restrictions do we need on this function's \( x \)-values? For division to be defined, we'll need the denominator to be non-zero.

\[ \begin{align} &x^2 - 4 \neq 0 \end{align} \]

Let's see what \( x \)-values must be disallowed.

\[ \begin{align} &(x + 2) \mult (x - 2) \neq 0 \\ &x \neq -2 \andSpaced x \neq 2 \end{align} \]

All inputs except \( x = -2 \) and \( x = 2 \) may be safely allowed.

Apart from division, there is only one other component that needs a restriction at this time.

As we saw in the preceding section, the square root requires its input to be positive.

Example.

Let's see what restrictions we need for the function

\[ \begin{align} z &\depends x \\ z &= \sqrt[2]{x^2 + 3} - 5. \end{align} \]

In order for our function to be defined, we'll need the input to the square root to be positive.

\[ \begin{align} &x^2 + 3 \ge 0 \end{align} \]

But wait a second! When we square a number \( x \), the result, \( x^2 \), is always positive. And adding three will only make it larger. Since every \( x \)-value will make \( x^2 + 3 \) positive, we don't need any restrictions on \( x \) for this function.

We'll say that a function is undefined at each of its disallowed inputs. So in particular,

\[ \begin{align} &\frac{5}{0} & &\andSpaced & &\sqrt[2]{-4} \end{align} \]

are both undefined.

Division By Zero

As we study Calculus, there will be times when we will be tempted to divide by zero. Let's take a moment to think through why this is a bad idea.

Question.

What's so wrong with dividing by zero?

For the sake of argument, suppose that we defined the fraction \( 1 / 0 \) to be some number. Then we could write

\[ \begin{align} 0 \mult \frac{1}{0} &= 0 & &\andSpaced & 0 \mult \frac{1}{0} &= 1. \end{align} \]

The first equality must hold because multiplying zero and any number gives zero. The second equality must hold because multiplying any number with its reciprocal gives one. We are forced to conclude that zero and one are equal, but this collapses our mathematical world! We must therefore be vigilant: our computations can never include a division by zero.