The definitional domain of a function

The defining scope of a function is all the allowable values that we can put after the argument x in the function f(x) so that the function makes sense.

What is the definitional scope of a function

A simple example is the function f(x) = 1/x. The definitional domain is the set of all admissible values that, when substituted into the function 1/x, yield a valid expression.

To know what a "valid expression" is, we need to know the properties of functions and operations. In our function, we divide the variable x, and the variable x is in the denominator of the fraction. What values can we divide by? All but zero. We can't divide by zero, nor can zero be in the denominator of a fraction. There is no other restriction on division.

Therefore, we can put any real number after x except zero. The definitional domain is thus equal: D(f) = ℝ ∖ {0}. We usually write the defining domain using the letter D and write the function whose defining domain we are calculating in parentheses.

At this point, we have calculated the largest definitional domain. But since we have defined the defining scope as the set of values that we can place after x, any subset of the defining scope will in turn be the defining scope. Sometimes we may find it useful to restrict a function to only some x. For example, in sequences we restrict ourselves to natural numbers, in goniometry we often count only the interval <0, 2π>.

However, if we are given an example where the defining range of a function is to be calculated, you are usually expected to calculate the largest possible defining range.

We can also read the defining range from the graph of the function. For an example, take the graph of the previous function f(x) = 1/x.

If you project the graph onto the axis of x, you get the definitional domain. If the point x is not an element of the defining domain, then if you make a vertical perpendicular to the axis x at that point , that line will not intersect any point on the graph. We see that only zero satisfies this condition - it has no point "above or below" any point on the graph of the function. All the others do.

Calculating the defining domain

To calculate the defining domain of a composite function, you need to know the defining domains of all the functions that make up the composite function. Otherwise, you cannot. If I were to ask you to calculate the definitional scope of the function f(x) = raz(x) + dva(x), you would fail because you don't know how the functions raz and dva are defined. Let's show the definitional scopes of some elementary functions:

• Thelogarithm has D(f) = (0, ∞).
• Thesquare root has D(f) = <0, ∞).
• Tangent has D(f) = ℝ ∖ {π/2 + Kπ}; K ∈ ℤ.
• Thecotangent has D(f) = ℝ ∖ {Kπ}; K ∈ ℤ.
• Theexponential function has D(f) = ℝ.
• Linear functions and quadratic functions have D(f) = ℝ.

If you are given only one function, it is very simple. Simply find out what definitional domain the function has, look at the argument, and then just plug it into the (in)equation as appropriate. So example - determine the definitional domain of the function $\log (3x + 2)$. The argument of the logarithm must be positive, so 3x + 2 > 0 must hold. This is a linear inequality that we can easily calculate:

$$\begin{eqnarray} 3x + 2 &>& 0\\ 3x&>&-2\\ x&>&-\frac23 \end{eqnarray}$$

The defining domain is all x that satisfy this inequality, so D(f) = (−2/3, ∞).

Calculate the defining domain of $\sqrt{2x + 8}$.

There must not be a negative number below the square root, so we solve the inequality 2x + 8 ≥ 0, giving the result:

$$\begin{eqnarray} 2x+8 &\ge& 0\\ 2x&\ge&-8\\ x&\ge&-4 \end{eqnarray}$$

So D(f) = <−4, ∞).

Calculate the defining domain 1/x².

There must not be a zero in the denominator, so we solve the equation x² ≠ 0. This is the case when x ≠ 0, so D(f) = ℝ ∖ {0}.

The decomposition of the composite functions

For composite functions, the situation is slightly more complicated, because we have to take into account multiple functions that interact with each other and gradually reduce their definitional domain. Take this example:

$$f(x) = \frac{1}{\sqrt{x}}$$

Here two conditions must hold. The term under the square root must be greater than or equal to zero, and it must not be negative. Therefore x ≥ 0. But the square root is in the denominator of the fraction, so it must be true that the square root must not come out to zero, then again the fraction would be meaningless. So we amplify the previous statement at x > 0.

We'll do a similar thing for the other examples. The important thing, though, is to determine the sequence of nested functions -- we have to figure out which function is inner and which function is outer. In the previous example, this was obvious -- the inner function was the square root and the outer function was the fraction. If we wanted to compute the expression by hand, we would first square the argument x and then compute the fraction. That's a pretty good guide -- the innermost function is the one you'd start with if you were calculating it by hand. Another example:

$$f(x) = \sqrt[5]{(\ln(\mbox{tan}(x)))^3}$$

We can now take the opposite approach. We'll look for the outer function. The outer function "wraps" the inner function. Here, quite obviously, the fifth root wraps all the remaining functions. Next, the third power wraps the result of the logarithm, the logarithm wraps the tangent, and finally the variable x wraps the tangent. Simple. The older method can be taken as follows: knowing the value of x, we first apply it to the tangent. So it's the innermost function. We apply this result to the logarithm -- so it's the second innermost function. We again apply this result to the power, and then square it off at the end.

Decompose the composite function:

$$f(x) = \sin \frac{42}{\ln(10^{2x+3})}$$

We start with the outer function. This is the sine. Next in order is the fraction, then the logarithm, then the exponential function (imagine this substitution: 10^a), and finally the linear function 2x + 3.

Decompose the composite function:

$$\Large f(x) = 2^{\frac{1}{\sqrt{3x}}}$$

The outer function is the exponential function 2^a. Next is the fraction, then in the denominator of the fraction is the square root, and finally the linear function 3x.

The definitional domain of composite functions

If you know how to decompose functions into outer and inner functions, it can be no problem for you to determine the defining domain. Let's demonstrate this right away with an example:

$$f(x) = \frac{47}{\sin(x)-1}$$

Start with a fraction. For this fraction, the denominator must be different from zero. Let's write it down:

$$\begin{eqnarray} \sin(x)-1 &\ne& 0\\ \sin(x) &\ne&+1 \end{eqnarray}$$

From the properties of the sine function, we know that it reaches the value 1 at the point x = π/2. Since it is a periodic function, we get the set K = {π/2 + 2Kπ}; K ∈ ℤ. This is the set of values for which the denominator comes out equal to zero. We cannot plug this x into our function.

But that's not all, we still need to look at the definitional domain of the sine function. Fortunately, this is equal to ℝ, so it doesn't limit us in any way. The restriction is only due to the denominator of the fraction. So D(f) = ℝ ∖ {π/2 + 2Kπ}; K ∈ ℤ.

For more solved examples, see MatWiki's Definitional Scopes of Functions category.