Functions H
The Skill
A function is a rule that takes an input, does something to it, and produces exactly one output.
Function Notation
Instead of writing $y = 2x + 3$, we write: $$f(x) = 2x + 3$$
This reads as "f of x equals 2x plus 3".
$f(x)$ means: Apply the function $f$ to the input $x$.
Evaluating Functions
To find $f(5)$, substitute $x = 5$: $$f(x) = 2x + 3$$ $$f(5) = 2(5) + 3 = 13$$
We can also use different inputs:
- $f(-2) = 2(-2) + 3 = -1$
- $f(a) = 2a + 3$
- $f(x+1) = 2(x+1) + 3 = 2x + 5$
Composite Functions
A composite function applies one function, then another.
If $f(x) = 2x + 1$ and $g(x) = x^2$:
$$fg(x) = f(g(x))$$
This means: First apply $g$, then apply $f$ to the result.
$$fg(x) = f(g(x)) = f(x^2) = 2(x^2) + 1 = 2x^2 + 1$$
Warning: $fg(x) \neq gf(x)$ in general!
$$gf(x) = g(f(x)) = g(2x+1) = (2x+1)^2 = 4x^2 + 4x + 1$$
Memory tip: Read $fg(x)$ from right to left — start with $g$, then $f$.
Inverse Functions
The inverse function $f^{-1}(x)$ reverses what $f$ does.
If $f(x) = 2x + 3$, then $f^{-1}(x)$ undoes this.
Finding the Inverse
- Write $y = f(x)$
- Swap $x$ and $y$
- Rearrange to make $y$ the subject
- Replace $y$ with $f^{-1}(x)$
Example: Find the inverse of $f(x) = 2x + 3$
Step 1: $y = 2x + 3$ Step 2: $x = 2y + 3$ Step 3: $x - 3 = 2y$, so $y = \frac{x-3}{2}$ Step 4: $f^{-1}(x) = \frac{x-3}{2}$
Properties of Inverses
- $f(f^{-1}(x)) = x$
- $f^{-1}(f(x)) = x$
- The graph of $f^{-1}$ is a reflection of $f$ in the line $y = x$
Domain and Range
Domain: The set of allowed inputs (x-values). Range: The set of possible outputs (y-values).
For example, if $f(x) = \sqrt{x}$:
- Domain: $x \geq 0$ (can't square root negatives)
- Range: $f(x) \geq 0$ (square roots are non-negative)
The Traps
Common misconceptions and how to avoid them.
Getting the order wrong in composite functions "The Order Flip"
The Mistake in Action
Given $f(x) = x + 3$ and $g(x) = 2x$. Find $fg(4)$.
Wrong: $fg(4) = f(4) \times g(4) = 7 \times 8 = 56$
Why It Happens
Students interpret $fg(x)$ as "f times g" rather than "f of g" — a composition.
The Fix
$fg(x)$ means $f(g(x))$ — apply $g$ first, then $f$.
Read from right to left: $fg(4)$ = f of (g of 4)
Step 1: Find $g(4)$ $$g(4) = 2 \times 4 = 8$$
Step 2: Find $f(8)$ $$f(8) = 8 + 3 = 11$$
So $fg(4) = 11$
Not: $f(4) \times g(4)$ — that would be written as $f(4) \cdot g(4)$
Spot the Mistake
$fg(4)$
$= f(4) \times g(4) = 56$
Click on the line that contains the error.
Thinking inverse means reciprocal "The Reciprocal Trap"
The Mistake in Action
Given $f(x) = 2x + 3$. Find $f^{-1}(x)$.
Wrong: $f^{-1}(x) = \frac{1}{2x + 3}$
Why It Happens
Students confuse $f^{-1}(x)$ (inverse function) with $\frac{1}{f(x)}$ (reciprocal) because of the negative exponent notation.
The Fix
$f^{-1}(x)$ is NOT $\frac{1}{f(x)}$!
The inverse function reverses what $f$ does. If $f$ takes 1 to 5, then $f^{-1}$ takes 5 back to 1.
To find $f^{-1}(x)$:
- Let $y = 2x + 3$
- Swap: $x = 2y + 3$
- Rearrange: $x - 3 = 2y$, so $y = \frac{x - 3}{2}$
- Therefore: $f^{-1}(x) = \frac{x - 3}{2}$
Check: $f(1) = 5$ and $f^{-1}(5) = \frac{5-3}{2} = 1$ ✓
Spot the Mistake
Find $f^{-1}(x)$ where $f(x) = 2x + 3$
$f^{-1}(x) = \frac{1}{2x + 3}$
Click on the line that contains the error.
Not substituting properly into function expressions "The Substitution Slip"
The Mistake in Action
Given $f(x) = x^2 + 1$. Find $f(x + 2)$.
Wrong: $f(x + 2) = x^2 + 1 + 2 = x^2 + 3$
Why It Happens
Students add 2 to the expression instead of replacing every $x$ with $(x + 2)$.
The Fix
$f(x + 2)$ means replace every $x$ in the formula with $(x + 2)$.
$f(x) = x^2 + 1$
$f(x + 2) = (x + 2)^2 + 1$ $= x^2 + 4x + 4 + 1$ $= x^2 + 4x + 5$
The pattern: $f(\text{something})$ means put that "something" wherever you see $x$ in the formula.
Spot the Mistake
$f(x) = x^2 + 1$, find $f(x+2)$
$f(x + 2) = x^2 + 1 + 2 = x^2 + 3$
Click on the line that contains the error.
Assuming fg(x) equals gf(x) "The Commutative Confusion"
The Mistake in Action
Given $f(x) = x^2$ and $g(x) = x + 1$.
Student states: "$fg(x) = gf(x)$ because multiplication is commutative."
Why It Happens
Students apply the commutative property of multiplication to function composition, which doesn't work.
The Fix
Function composition is NOT commutative: $fg(x) \neq gf(x)$ in general.
$fg(x) = f(g(x)) = f(x + 1) = (x + 1)^2 = x^2 + 2x + 1$
$gf(x) = g(f(x)) = g(x^2) = x^2 + 1$
These are different! The order matters because different operations are being done.
Think of it like getting dressed: Socks then shoes ≠ Shoes then socks!
Spot the Mistake
$fg(x) = gf(x)$ because multiplication is commutative
Click on the line that contains the error.
The Deep Dive
Apply your knowledge with these exam-style problems.
Level 1: Fully Worked
Complete solutions with commentary on each step.
Question
Given $f(x) = 3x² - 2x + 1$, find: a) $f(4)$ b) $f(-2)$ c) $f(a)$
Solution
a) f(4): Replace every $x$ with 4. $$f(4) = 3(4)² - 2(4) + 1$$ $$= 3(16) - 8 + 1$$ $$= 48 - 8 + 1 = 41$$
b) f(-2): Replace every $x$ with -2. $$f(-2) = 3(-2)² - 2(-2) + 1$$ $$= 3(4) - (-4) + 1$$ $$= 12 + 4 + 1 = 17$$
c) f(a): Replace every $x$ with $a$. $$f(a) = 3a² - 2a + 1$$
Answers: a) 41, b) 17, c) $3a² - 2a + 1$
Question
Given $f(x) = 2x + 1$ and $g(x) = x²$, find: a) $fg(3)$ b) $gf(3)$
Solution
a) fg(3): Apply g first, then f.
Step 1: $g(3) = 3² = 9$ Step 2: $f(9) = 2(9) + 1 = 19$
So $fg(3) = 19$
b) gf(3): Apply f first, then g.
Step 1: $f(3) = 2(3) + 1 = 7$ Step 2: $g(7) = 7² = 49$
So $gf(3) = 49$
Notice: $fg(3) ≠ gf(3)$ — order matters!
Answers: a) 19, b) 49
Question
Find $f^{-1}(x)$ when $f(x) = \frac{x - 3}{4}$
Solution
Step 1: Write $y = f(x)$ $$y = \frac{x - 3}{4}$$
Step 2: Swap $x$ and $y$ $$x = \frac{y - 3}{4}$$
Step 3: Rearrange to make $y$ the subject. Multiply both sides by 4: $$4x = y - 3$$
Add 3 to both sides: $$4x + 3 = y$$
Step 4: Write the inverse. $$f^{-1}(x) = 4x + 3$$
Check: $f(7) = \frac{7-3}{4} = 1$ and $f^{-1}(1) = 4(1) + 3 = 7$ ✓
Answer: $f^{-1}(x) = 4x + 3$
Level 2: Scaffolded
Fill in the key steps.
Question
Given $f(x) = x + 5$ and $g(x) = 3x$, find $fg(x)$.
Level 3: Solo
Try it yourself!
Question
$f(x) = x² + 2$ where $x ≥ 0$. Find $f^{-1}(x)$ and state its domain.
Show Solution
Step 1: Write $y = x² + 2$
Step 2: Swap: $x = y² + 2$
Step 3: Rearrange for $y$: $$x - 2 = y²$$ $$y = ±\sqrt{x - 2}$$
Step 4: Choose the correct root. Since the original function has domain $x ≥ 0$, the outputs are $f(x) ≥ 2$. The inverse should give non-negative outputs (matching the original domain). So we take the positive root.
$$f^{-1}(x) = \sqrt{x - 2}$$
Domain of inverse: $x ≥ 2$ (the range of the original function)
Answers: $f^{-1}(x) = \sqrt{x - 2}$, domain: $x ≥ 2$
Question
Given $f(x) = 2x - 3$, solve $ff(x) = 5$.
Show Solution
Step 1: Find $ff(x)$. $$ff(x) = f(f(x)) = f(2x - 3)$$ $$= 2(2x - 3) - 3$$ $$= 4x - 6 - 3$$ $$= 4x - 9$$
Step 2: Solve $ff(x) = 5$. $$4x - 9 = 5$$ $$4x = 14$$ $$x = \frac{14}{4} = \frac{7}{2} = 3.5$$
Check: $f(3.5) = 2(3.5) - 3 = 4$ $ff(3.5) = f(4) = 2(4) - 3 = 5$ ✓
Answer: $x = 3.5$ (or $\frac{7}{2}$)
Examiner's View
Mark allocation: Evaluating functions is 1-2 marks. Composite functions are 2-3 marks. Finding inverses is 3-4 marks. Domain/range questions are 2-3 marks.
Common errors examiners see:
- Getting the order wrong in composite functions ($fg \neq gf$)
- Not substituting properly when finding $f(x+1)$ or similar
- Forgetting to swap $x$ and $y$ when finding inverses
- Incorrect rearranging when finding inverses
- Using $f^{-1}(x) = \frac{1}{f(x)}$ (this is WRONG)
What gains marks:
- Showing substitution clearly
- For composites: showing the intermediate step
- For inverses: showing all four steps clearly
- Checking: $f(f^{-1}(3))$ should equal 3
AQA Notes
AQA often asks you to find $fg(x)$ then evaluate at a specific value, like $fg(2)$.