Relations and Functions • Topic 3 of 3

Functions and Standard Real Functions

A function $f$ from $A$ to $B$ is a relation in which every element of $A$ is paired with exactly one element of $B$. Two demands, both essential: every input is used (no element of $A$ left out), and each input gives one output (never two). We write $f : A \to B$ and $y = f(x)$.

$$f : A \to B \text{ is a function} \iff \text{each } x \in A \text{ has a unique image } f(x) \in B$$

The vertical line test captures this graphically: a curve is a function exactly when no vertical line meets it more than once (one $x$, one $y$).

Standard real functions you must recognise by formula and graph:

FunctionRuleDomainRange
Identity$f(x) = x$$\mathbb{R}$$\mathbb{R}$
Constant$f(x) = c$$\mathbb{R}$$\{c\}$
Modulus$f(x) = |x|$$\mathbb{R}$$[0, \infty)$
Signum$f(x) = \dfrac{|x|}{x}\,(x\ne0)$$\mathbb{R}$$\{-1, 0, 1\}$
Greatest integer$f(x) = [x]$$\mathbb{R}$$\mathbb{Z}$
Reciprocal$f(x) = \dfrac{1}{x}$$\mathbb{R}\setminus\{0\}$$\mathbb{R}\setminus\{0\}$

Algebra of real functions: for functions with overlapping domains you may add, subtract, multiply and divide them pointwise: $(f \pm g)(x) = f(x) \pm g(x)$, $(fg)(x) = f(x)g(x)$, and $\left(\tfrac{f}{g}\right)(x) = \tfrac{f(x)}{g(x)}$ wherever $g(x) \ne 0$.

Finding the domain of a real function means listing every $x$ for which the rule gives a real value — exclude inputs that cause division by zero or an even root of a negative number. The range is the resulting set of outputs.

Deeper Insight — "exactly one output" is the entire idea, and it is what makes calculus possible: A function is a relation that has been disciplined by a single rule: one input, one output. That rule looks modest but it is the reason functions, not arbitrary relations, became the central object of mathematics — because a unique output per input is exactly what lets us speak of "the value of $f$ at $x$", differentiate it, integrate it, and invert it. The vertical line test is just this rule drawn on a graph. The standard functions in the table are worth memorising not as trivia but as a vocabulary: nearly every function you meet later is built by combining, transforming or composing these few, so knowing their shapes (the V of $|x|$, the steps of $[x]$, the three-level jump of the signum) lets you predict the behaviour of complicated expressions at a glance. Master "unique image" and the standard graphs, and the leap to Limits and Derivatives in Chapter 12 becomes a short step rather than a cliff.

Modulus and greatest-integer function graphs f(x) = |x|xyV-shape, range [0, ∞) f(x) = [x] (step)jumps at each integer, range ℤ
1
Worked Example
Is $R = \{(1,2),(2,3),(1,4)\}$ from $A=\{1,2\}$ a function? Why or why not?
Solution
  1. Check each input has a unique output.
  2. Input $1$ maps to both $2$ and $4$ — two images.

Answer: No — it is a relation but not a function, since $1$ does not have a unique image.

2
Worked Example
Find the domain of $f(x) = \dfrac{1}{x^2 - 4}$.
Solution
  1. The denominator must be non-zero: $x^2 - 4 \ne 0$.
  2. $x^2 \ne 4 \Rightarrow x \ne \pm 2$.

Answer: Domain $= \mathbb{R} \setminus \{-2, 2\}$.

3
Worked Example
Find the domain and range of $f(x) = \sqrt{x - 3}$.
Solution
  1. The radicand must be $\ge 0$: $x - 3 \ge 0 \Rightarrow x \ge 3$.
  2. The square root yields values $\ge 0$.

Answer: Domain $= [3, \infty)$; Range $= [0, \infty)$.

4
Worked Example
Evaluate the greatest integer function: $[2.7]$, $[-2.7]$ and $[5]$.
Solution
  1. $[x]$ is the greatest integer $\le x$.
  2. $[2.7] = 2$.
  3. $[-2.7] = -3$ (since $-3 \le -2.7 < -2$).
  4. $[5] = 5$.

Answer: $[2.7] = 2$, $[-2.7] = -3$, $[5] = 5$.

5
Worked Example
If $f(x) = x^2$ and $g(x) = 2x + 1$, find $(f + g)(x)$ and $(fg)(2)$.
Solution
  1. $(f+g)(x) = f(x) + g(x) = x^2 + 2x + 1$.
  2. $(fg)(2) = f(2) \cdot g(2) = (2^2)(2\cdot2+1) = 4 \times 5 = 20$.

Answer: $(f+g)(x) = x^2 + 2x + 1$; $(fg)(2) = 20$.

6
Worked Example
Find the range of $f(x) = |x - 2| + 1$.
Solution
  1. $|x-2| \ge 0$ for all $x$, with minimum $0$ at $x = 2$.
  2. So $f(x) = |x-2| + 1 \ge 0 + 1 = 1$, and it grows without bound.

Answer: Range $= [1, \infty)$.

Key Points

  • A function $f:A\to B$ assigns to every $x\in A$ exactly one image $f(x)\in B$.
  • Vertical line test: a graph is a function if no vertical line cuts it more than once.
  • Know the standard functions and graphs: identity, constant, $|x|$, signum, $[x]$, $1/x$.
  • Domain excludes division by zero and even roots of negatives; functions add/subtract/multiply/divide pointwise.
Tap an option to check your answer0 / 4
Q1.A function from $A$ to $B$ assigns to each element of $A$:
Explanation: A function gives each input exactly one output.
Q2.The domain of $f(x)=\dfrac{1}{x-2}$ is:
Explanation: The denominator must be non-zero.
Q3.The range of $f(x)=x^2$ (with $f:\mathbb{R}\to\mathbb{R}$) is:
Explanation: Squares are non-negative.
Q4.The domain of $f(x)=\sqrt{x}$ is:
Explanation: The radicand must be non-negative.