Domain and range of real functions

In Class XI, when you write $f(x) = \dfrac{\sqrt{x - 1}}{x - 3}$ without specifying a domain, the convention is that the domain is the largest subset of $\mathbb{R}$ for which the formula makes sense. We call this the natural domain. Computing it is a step-by-step affair; computing the range is the harder half.

Definitions

Given a formula $f(x)$, the natural domain is $\text{Dom}(f) = \{x \in \mathbb{R} : f(x) \text{ is a real number}\}.$

The range is $\text{Range}(f) = \{f(x) : x \in \text{Dom}(f)\}.$

Equivalently, $y$ is in the range iff the equation $f(x) = y$ has a real solution $x$ in the domain.

Restrictions to watch for

When you build the natural domain you must rule out:

1. Division by zero. Denominators must be non-zero. For $\dfrac{1}{g(x)}$, require $g(x) \ne 0$.
2. Square roots (and even roots) of negatives. $\sqrt{g(x)}$ requires $g(x) \ge 0$.
3. Logarithms of non-positive numbers. $\log g(x)$ requires $g(x) > 0$ (later chapters).
4. Trigonometric undefined points. $\tan x$ and $\sec x$ require $\cos x \ne 0$; $\cot x$ and $\csc x$ require $\sin x \ne 0$.

To find the domain of a compound expression, take the intersection of every individual restriction.

Strategies for finding the range

Several methods, used as the problem demands.

(a) Solving $y = f(x)$ for $x$. Express $x$ in terms of $y$; the range is the set of $y$'s for which $x$ is real and lies in the domain.

(b) Completing the square. Works for any quadratic.

(c) Inspection. For $|x|, x^2, \sqrt{x}$ etc., recognise the standard ranges.

(d) Calculus (later). Find maxima and minima , but this waits until Chapter 12.

(e) Graph reading. Draw the graph; the projection on the $y$-axis is the range.

Worked examples

Example 1. Find the domain of $f(x) = \dfrac{1}{x^2 - 4}$.

Denominator zero when $x^2 - 4 = 0$, i.e. $x = \pm 2$. So domain $= \mathbb{R} \setminus \{-2, 2\}$.

Example 2. Find the domain of $f(x) = \sqrt{x - 2} + \sqrt{5 - x}$.

Need $x - 2 \ge 0$ and $5 - x \ge 0$, i.e. $x \ge 2$ and $x \le 5$. Intersection: $[2, 5]$. Domain $= [2, 5]$.

Example 3. Find the range of $f(x) = x^2 - 6x + 11$.

Complete the square: $f(x) = (x - 3)^2 + 2 \ge 2$. Equality at $x = 3$. Domain is $\mathbb{R}$, so all values $\ge 2$ are attained. Range $= [2, \infty)$.

Example 4. Find the range of $f(x) = \dfrac{x}{1 + x^2}$.

Let $y = \dfrac{x}{1 + x^2}$. Then $y(1 + x^2) = x$, so $yx^2 - x + y = 0$. For real $x$ we need discriminant $\ge 0$: $1 - 4y^2 \ge 0 \iff y^2 \le \tfrac{1}{4} \iff -\tfrac{1}{2} \le y \le \tfrac{1}{2}.$ Range $= \left[-\tfrac{1}{2}, \tfrac{1}{2}\right]$. (Both endpoints achieved at $x = \pm 1$.)

Example 5 (harder). Find the domain and range of $f(x) = \dfrac{\sqrt{x - 1}}{x - 3}$.

Domain: need $x - 1 \ge 0$ and $x - 3 \ne 0$, i.e. $x \ge 1$ and $x \ne 3$. Domain $= [1, 3) \cup (3, \infty)$.

Range: write $y = \dfrac{\sqrt{x - 1}}{x - 3}$. Let $t = \sqrt{x - 1} \ge 0$, so $x = t^2 + 1$ and $x - 3 = t^2 - 2$. Then $y = \dfrac{t}{t^2 - 2}$ for $t \ge 0$, $t \ne \sqrt{2}$.

At $t = 0$: $y = 0$.

As $t \to \sqrt{2}^-$: $y \to -\infty$. As $t \to \sqrt{2}^+$: $y \to +\infty$. As $t \to \infty$: $y \to 0^+$.

Solving $y(t^2 - 2) = t \Rightarrow y t^2 - t - 2y = 0$, real $t$ exists iff discriminant $1 + 8y^2 \ge 0$, which is always true. But we also need $t \ge 0$.

A cleaner answer: range $= \mathbb{R}$. (You can verify: for any $y \in \mathbb{R}$, set $t = \dfrac{1 + \sqrt{1 + 8y^2}}{2y}$ when $y \ne 0$; check $t \ge 0$.)

Try it yourself

1. Find the domain of $f(x) = \dfrac{1}{x - 1}$.
2. Find the domain of $f(x) = \sqrt{16 - x^2}$.
3. Find the domain of $f(x) = \sqrt{x^2 - 4}$.
4. Find the range of $f(x) = x^2 + 2x + 5$.
5. Find the range of $f(x) = -x^2 + 6x - 8$.
6. Find the domain of $f(x) = \dfrac{1}{\sqrt{x - 1}}$.
7. Find the range of $f(x) = |x - 3|$.
8. Find the range of $f(x) = \dfrac{1}{1 + x^2}$.
9. Find the domain of $f(x) = \dfrac{x - 2}{x^2 - 5x + 6}$.
10. Find the range of $f(x) = \dfrac{x^2}{1 + x^2}$.
11. Find the domain of $f(x) = \sqrt{x - 1} + \sqrt{6 - x}$.
12. Find the range of $f(x) = \sin^2 x + \cos^4 x$ (use $\cos^2 x = 1 - \sin^2 x$).

Pitfalls / Tricks

• Always check all restrictions and intersect. A common error: $\sqrt{x - 1}/(x - 3)$ , students remember $x \ge 1$ but forget $x \ne 3$.
• For square roots, strict inequality $> 0$ is needed when the square root sits in a denominator.
• Range $\ne$ codomain. The codomain is declared; the range is computed.
• Insight. When the formula resists direct inspection, set $y = f(x)$ and solve for $x$. The conditions for $x$ to be real and inside the domain give the range exactly.