Worked examples · Sheet 1

Natural domains of functions

Theory used: Natural domain Intervals Sets and operations

Exercise 1

Find the natural domain of

f(x) = \frac{x + 2}{x^2 - 9} .

x^2 - 9 \neq 0 \iff (x - 3)(x + 3) \neq 0 \iff x \neq 3 \ \text{ and } \ x \neq -3

D = \mathbb{R} \setminus \{-3, 3\} = (-\infty, -3) \cup (-3, 3) \cup (3, +\infty)

The function is a quotient of polynomials, so the only condition is that the denominator be different from zero. The denominator is a difference of squares and factors as $(x - 3)(x + 3)$ ; a product is zero when at least one of the factors is, therefore the denominator vanishes only at $x = 3$ and at $x = -3$ . The numerator imposes no conditions, so at $x = -2$ the function is defined and equals $0$ . The domain is therefore $\mathbb{R}$ without the two points $-3$ and $3$ , which written with intervals is the union of three open intervals.

Exercise 2

Find the natural domain of

f(x) = \sqrt{\frac{x - 1}{x + 4}} .

\frac{x - 1}{x + 4} \geq 0

	$x < -4$	$-4 < x < 1$	$x > 1$
$x - 1$	$-$	$-$	$+$
$x + 4$	$-$	$+$	$+$
quotient	$+$	$-$	$+$

D = (-\infty, -4) \cup [1, +\infty)

The root has even index, so its argument must be greater than or equal to zero; moreover the argument is a fraction, and its denominator must be different from zero. The two conditions are satisfied together by studying the sign of the quotient. The numerator is positive for $x > 1$ , the denominator for $x > -4$ , and the quotient is positive where the two factors have the same sign, that is, for $x < -4$ and for $x > 1$ . The quotient equals zero at $x = 1$ , which therefore belongs to the domain, since $\sqrt{0} = 0$ . The point $x = -4$ makes the denominator zero and is excluded. The domain is therefore $(-\infty, -4) \cup [1, +\infty)$ .

Exercise 3

Find the natural domain of

f(x) = \ln\left(x^2 - 4x\right) + \frac{1}{\sqrt{6 - x}} .

\begin{cases} x^2 - 4x > 0 \\[4pt] 6 - x > 0 \end{cases} \iff \begin{cases} x < 0 \ \text{ or } \ x > 4 \\[4pt] x < 6 \end{cases}

D = (-\infty, 0) \cup (4, 6)

Above the axis the solutions of

x^{2} - 4 x > 0

, below it those of

6 - x > 0

: the domain is where the two strips overlap,

(- \infty, 0) \cup (4, 6)

The function is the sum of two terms, and each imposes its own conditions, which must hold together. The argument of the logarithm must be strictly positive. The polynomial $x^2 - 4x = x(x - 4)$ has as its graph an upward-opening parabola that vanishes at $0$ and at $4$ , so it is positive outside the interval $[0, 4]$ . In the second term the root is in the denominator, therefore its argument must be strictly positive, and not just greater than or equal to zero, because $\sqrt{0} = 0$ would make the denominator zero; this gives $x < 6$ . The domain is the intersection of the solutions of the two conditions, that is, $(-\infty, 0) \cup (4, 6)$ , as the diagram shows.