Applying Function Notation
Master function notation for Queensland Year 11 Mathematical Methods (QCAA). Writing a rule as f of x names the output for an input x — a compact way to describe any function.
You will learn to evaluate f of a for numbers and expressions, solve f of x equals a given value, read values from a graph, and find an unknown rule — everyday algebra used throughout the course.
Every question with a fully worked solution.
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Theory
In Year 11 Mathematical Methods (QCAA, Unit 1) function notation \(f(x)\) names the output of a rule for an input \(x\). This page shows how to evaluate \(f(a)\) (including negative inputs and expressions like \(f(a+1)\)), how to solve \(f(x)=k\), how to read a value off a graph, and how to find an unknown rule from given conditions.
Function notation \(f(x)\) reads “\(f\) of \(x\)” and stands for the output of the function when the input is \(x\). To evaluate \(f(a)\), substitute \(a\) for every \(x\) in the rule and simplify — use brackets so a negative input is handled correctly.
You can substitute an expression as the input: \(f(a+1)\) means replace every \(x\) with \((a+1)\), then expand. Solving \(f(x)=k\) reverses the process — set the rule equal to \(k\) and solve the resulting equation for \(x\).
A statement such as \(f(a)=b\) says “the input \(a\) gives the output \(b\)”, i.e. the point \((a,b)\) is on the graph. Given two such conditions you can determine an unknown rule, for example the gradient and intercept of a linear function.
Evaluating and solving with a rule \(f\):
A linear rule from its gradient and a point:
How to work with function notation
- Evaluate: to find \(f(a)\), replace every \(x\) in the rule with \(a\) in brackets, then simplify.
- Solve: to solve \(f(x)=k\), set the rule equal to \(k\) and solve the equation (factorise a quadratic, isolate \(x\) for a linear).
- Determine a rule: turn each condition \(f(a)=b\) into an equation in the unknown coefficients, then solve the system.
Replace every \(x\) with \((-2)\), keeping brackets:
| \(f(-2)\) | \(=\) | \(3(-2)^2-2(-2)+1\) |
Work through the powers and products:
| \(=\) | \(3(4)-2(-2)+1\) | |
| \(=\) | \(12+4+1\) | |
| \(=\) | \(17\) |
\(f(-2)=17\).
Set the rule equal to \(5\):
| \(x^2-2x-3\) | \(=\) | \(5\) |
Move everything to one side:
| \(x^2-2x-8\) | \(=\) | \(0\) |
Factorise and solve:
| \((x-4)(x+2)\) | \(=\) | \(0\) |
| \(x\) | \(=\) | \(4\ \text{or}\ x=-2\) |
\(x=4\) or \(x=-2\).
Replace every \(x\) with \((a+1)\):
| \(f(a+1)\) | \(=\) | \((a+1)^2+3(a+1)\) |
Expand each part:
| \(=\) | \(a^2+2a+1+3a+3\) |
Collect like terms:
| \(=\) | \(a^2+5a+4\) |
\(f(a+1)=a^2+5a+4\).
Gradient — use the two points \((2,7)\) and \((-1,-2)\):
| \(m\) | \(=\) | \(\dfrac{7-(-2)}{2-(-1)}\) |
| \(=\) | \(\dfrac{9}{3}\) | |
| \(=\) | \(3\) |
Intercept — substitute \(f(2)=7\) into \(f(x)=3x+c\):
| \(7\) | \(=\) | \(3(2)+c\) |
| \(7\) | \(=\) | \(6+c\) |
| \(c\) | \(=\) | \(1\) |
Write the rule:
| \(f(x)\) | \(=\) | \(3x+1\) |
\(f(x)=3x+1\).
Common pitfalls
Frequently asked questions
What does f(x) mean?
It reads “\(f\) of \(x\)” and stands for the output of the function \(f\) when the input is \(x\). It is a name for the \(y\)-value, not \(f\) times \(x\).
How do you evaluate a function at a value?
Replace every \(x\) in the rule with the input, in brackets, then simplify. For \(f(x)=3x^2-2x+1\), \(f(-2)=3(-2)^2-2(-2)+1=17\).
How do you solve f(x)=k?
Set the rule equal to \(k\) and solve for \(x\): isolate \(x\) for a linear rule, or move everything to one side and factorise for a quadratic.
What does f(a)=b tell you?
That the input \(a\) produces the output \(b\), so the point \((a,b)\) lies on the graph of \(f\).
How do you find f(a+1)?
Substitute \((a+1)\) for every \(x\) in the rule and expand. For \(f(x)=x^2+3x\), \(f(a+1)=(a+1)^2+3(a+1)=a^2+5a+4\).
How do you find the rule of a linear function from two values?
Find the gradient \(m\) from the two points, then substitute one point into \(f(x)=mx+c\) to find \(c\).