Domain

Try to compute $\dfrac{1}{0}$ on a calculator, and then $\sqrt{-4}$. Both refuse. Say, for each, what the refusal actually means: is the answer very large, or is there no answer at all? The difference matters for how you describe the domain.

The domain of a function is the set of all allowable values of the input. Unless a question says otherwise, take it to be every real number that does not break the rule — the natural domain.

The domain of a function is the set of values the input is allowed to take.

When a question gives no restriction, the domain is understood to be the natural domain: every real number for which the rule produces a value. The method is always the same.

1. Assume the domain is all real numbers.
2. Ask what, if anything, would break.
3. Remove those inputs.

So $f(x) = 3x - 7$ has domain all real numbers: no input breaks it. So does $f(x) = x^2 + 5x$, and every polynomial from the previous focus area, since polynomials are built from multiplication and addition alone and those work for every real number.

Most functions you meet have domain all real numbers. The skill is recognising the two families that do not, and the next card names them.

A zero denominator. Division by zero is undefined, so any input making a denominator zero is excluded. Set the denominator equal to zero, solve, and remove the solutions.

$$f(x) = \frac{5}{x-2}: \quad x - 2 = 0 \ \Rightarrow \ x = 2 \ \text{is excluded}$$

The domain is every real number except $2$. Note that the numerator plays no part: $\dfrac{0}{x-2}$ would have the same exclusion.

A negative under a square root. No real number squares to a negative, so $\sqrt{\text{negative}}$ has no real value. Set what is under the root to be at least zero and solve the inequality.

$$g(x) = \sqrt{x-3}: \quad x - 3 \geq 0 \ \Rightarrow \ x \geq 3$$

Here the exclusion removes a whole half of the number line rather than a single point.

The $\geq$ rather than $>$ matters: $\sqrt{0} = 0$ is perfectly well defined, so the endpoint is included. A root permits zero; a denominator does not. That single difference is why one endpoint is filled and the other is hollow.

The same domain can be written three ways, and questions use all of them.

As an inequality or an exclusion. Plain and usually clearest:

$$x \geq 3 \qquad \text{or} \qquad x \neq 2$$

In set-builder notation. The colon reads as "such that":

$$\{x : x \geq 3\} \qquad \{x : x \neq 2\}$$

In interval notation. A square bracket includes the endpoint, a round bracket excludes it, and $\infty$ always takes a round bracket because it is not a number that can be reached:

$$[3, \infty) \qquad (-\infty, 2) \cup (2, \infty)$$

That last one reads "everything below $2$, together with everything above $2$", and the $\cup$ symbol means the two pieces are taken together. A domain with a single point removed always needs two intervals, because removing an interior point splits the line in two.

A few more examples: $-1 < x \leq 4$ is $(-1, 4]$; all real numbers is $(-\infty, \infty)$; and $x > 0$ is $(0, \infty)$.

The natural domain is the largest sensible set of inputs for the algebra. A real situation often allows fewer.

Suppose a square has side length $x$ metres and area $A(x) = x^2$. As algebra, the natural domain is all real numbers: $(-2)^2$ is a perfectly good number. As a description of a square, a negative side length is meaningless, so the domain should be restricted to $x > 0$.

Similarly, if $C(n) = 4n + 20$ gives the cost of $n$ tickets, then $n$ must be a whole number and cannot be negative. The domain is $n = 0, 1, 2, \ldots$, and the graph is a row of isolated points, not a continuous line.

Two questions decide the restriction:

Can the quantity be negative? Can it take values between whole numbers?

Answering both, and saying so explicitly, is what a full answer to a modelling question requires. Stating a domain is part of defining the function, not an optional extra.

Five items on domains. Work each one, then reveal the answer.

1. 1 State the natural domain of $f(x) = 4x - 9$.

   Show answer
   Nothing breaks: no denominator and no root.All real numbers

2. 2 State the natural domain of $f(x) = \dfrac{3}{x+5}$.

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   Solve $x + 5 = 0$ and exclude the answer.$x \neq -5$

3. 3 State the natural domain of $f(x) = \sqrt{x+2}$.

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   Solve $x + 2 \geq 0$.$x \geq -2$

4. 4 Write $-3 \leq x < 8$ in interval notation.

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   Square bracket includes, round bracket excludes.$[-3, 8)$

5. 5 State the natural domain of $f(x) = x^2 - 7x + 1$.

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   A polynomial accepts every real input.All real numbers