Cambridge O Level Additional Mathematics · Syllabus 4037 · Functions
Function
What is Function?
A rule that assigns exactly one output to every input in a stated set called the domain. Two different inputs are allowed to share an output, but a single input may never produce two different outputs; a rule that does so is a relation rather than a function.
This definition is part of the Functions chapter in Cambridge O Level Additional Mathematics.
Function in context
A function is a rule that gives exactly one output for every input it is allowed to take. Everything else in this topic is a consequence of that one sentence: the domain is the set of inputs you are allowed to use, the range is the set of outputs actually produced, a composite feeds one function's output into another, and an inverse undoes the rule — which is only possible when no two inputs share an output.
Common mistakes with Function
- “\(f^{2}(x)\) means \([f(x)]^{2}\).” Correct \(f^{2}(x)=f(f(x))\) — apply \(f\) twice. For \(f(x)=2x+1\): \(f^{2}(3)=f(7)=15\), while \([f(3)]^{2}=7^{2}=49\). The syllabus does not use this iteration notation with trigonometric functions.
- “Composites can be read left to right, like English.” Correct In \(fg(x)\) the function nearest \(x\) acts first, so \(g\) goes first and \(f\) second. \(fg\) and \(gf\) are different functions in general.
- “Every function has an inverse; I just have to find it.” Correct Only a one‑one function has an inverse function. \(f(x)=x^{2}\) on \(\mathbb{R}\) has none, because \(f(2)=f(-2)=4\). Restricting the domain to \(x\ge 0\) is what makes an inverse possible.
- “I found the rule for \(f^{-1}(x)\), so I know everything about it.” Correct The rule is half of the function; the domain is the other half, and it is not free to choose — the domain of \(f^{-1}\) is the range of \(f\). For \(f(x)=e^{2x}\) the rule \(f^{-1}(x)=\tfrac{1}{2}\ln x\) carries the domain \(x>0\). Find it every time, so that you can state it the moment a question asks.
Questions students ask about Function
What is the difference between the domain and the range of a function?
The domain is the set of inputs a function is allowed to take; the range (or image set) is the set of outputs that domain actually produces. The domain is stated or chosen first, and the range follows from it — change the domain and the range changes with it. For \(f(x)=x^{2}\) with domain \(x\ge 0\), the range is \(f(x)\ge 0\).
Does \(f^{-1}(x)\) mean \(\dfrac{1}{f(x)}\)?
No. The superscript \(-1\) in \(f^{-1}\) is not an index; it names the inverse function, the mapping that runs \(f\) backwards. If \(f(2)=7\) then \(f^{-1}(7)=2\), whereas \(\dfrac{1}{f(2)}=\dfrac{1}{7}\). In the same way \(f^{2}(x)\) means \(f(f(x))\), apply \(f\) twice, and not \([f(x)]^{2}\).
Why does a many-one function have no inverse?
An inverse must send each output back to exactly one input. If two different inputs share an output, the inverse would have to return two values for that one input, which a function cannot do. Use the horizontal line test: a horizontal line meeting the graph more than once shows the function is many-one, so \(f(x)=x^{2}\) on \(\mathbb{R}\) has no inverse because \(f(2)=f(-2)=4\). Restricting the domain to \(x\ge 0\) makes it one-one and an inverse exists.
In \(fg(x)\), which function is applied first?
The function nearest \(x\) acts first, so in \(fg(x)=f(g(x))\) you apply \(g\) first and then \(f\) to its output. Composites are read from the inside out, not left to right like English, and \(fg\) and \(gf\) are in general different functions. A value \(x\) is a legal input only if it is in the domain of \(g\) and \(g(x)\) is in the domain of \(f\).
How do you find the domain of an inverse function?
The domain of \(f^{-1}\) is the range of \(f\), and the range of \(f^{-1}\) is the domain of \(f\). So after rearranging to find the rule for \(f^{-1}(x)\), go back to the original function, find its range, and state that as the domain of the inverse. The rule is only half the answer; an inverse without its domain is incomplete, and the domain is not free to choose.
How do you show that a function and its inverse are reflections in \(y=x\)?
Reflecting a point in the line \(y=x\) swaps its coordinates, so the point \((a,b)\) on \(y=f(x)\) corresponds to \((b,a)\) on \(y=f^{-1}(x)\). In an exam, sketch \(y=f(x)\), draw the dashed line \(y=x\), and reflect the curve in it, making sure the domain and range have swapped over. Any point where the two graphs meet lies on the line \(y=x\) itself.