Functions and Algebraic Fractions (Higher)

This lesson is Higher-tier only — every topic here is examined on the Higher papers of OCR GCSE Mathematics (J560) and does not appear on Foundation, so if you are sitting the Foundation tier you can treat this lesson as enrichment rather than required study. It covers function notation $f(x)$ , evaluating functions, composite functions, inverse functions, and simplifying, adding and subtracting algebraic fractions. These ideas tie together much of the algebra you have already met: evaluating a function is substitution, finding an inverse is rearranging to change the subject, and handling algebraic fractions relies on the factorising skills from earlier lessons. The manipulation is AO1; combining functions and fractions correctly, and explaining the order of operations in a composite, draws on AO2 and AO3.

Key Vocabulary

Term	Meaning
Function	A rule that assigns exactly one output to each input
$f(x)$	" $f$ of $x$ " — the output of function $f$ for input $x$
Composite function	Applying one function then another, e.g. $fg(x)$ means do $g$ first
Inverse function $f^{-1}(x)$	The function that undoes $f$
Algebraic fraction	A fraction whose numerator or denominator contains a variable
Common denominator	A shared denominator used to add or subtract fractions
Domain	The set of allowed inputs to a function

A helpful way to picture a function is as a machine: you feed a number in, the machine applies its rule, and a single number comes out. The notation $f(x)$ is just a compact way of naming that machine ( $f$ ) and its input ( $x$ ). Once you are comfortable with this idea, composite functions (two machines in a row) and inverse functions (a machine that runs the first one backwards) follow naturally. Algebraic fractions, the second half of this lesson, draw on the factorising skills you met earlier in the course, now applied to the tops and bottoms of fractions.

Function Notation and Evaluating

A function $f$ takes an input and produces an output. The notation $f(x)$ means "apply the rule $f$ to $x$ ". To evaluate $f(3)$ , replace every $x$ in the rule with $3$ . Be careful: $f(x)$ does not mean $f$ multiplied by $x$ — the brackets here signal "the function $f$ applied to $x$ ", a completely different idea from multiplication. You will also meet several named functions in the same question, such as $f$ , $g$ and $h$ ; each has its own rule, so always check which function you are being asked to use. The letter inside the brackets is just a placeholder, so $f(a)$ , $f(2)$ and $f(x + 1)$ all mean "put whatever is in the brackets in place of $x$ ".

Worked Example 1

Given $f(x) = 3x - 2$ , work out $f(4)$ .

Replace $x$ with $4$ : $f(4) = 3(4) - 2 = 10$ .

Answer: $10$

Worked Example 2

Given $g(x) = x^{2} + 1$ , work out $g(-3)$ .

Replace $x$ with $-3$ : $g(-3) = (-3)^{2} + 1 = 9 + 1 = 10$ .

Answer: $10$

Common error: writing $(-3)^{2} = -9$ . Squaring a negative gives a positive, so $g(-3) = 10$ .

Worked Example 3 — solving with function notation

Given $f(x) = 2x + 7$ , solve $f(x) = 19$ .

Set $2x + 7 = 19$ : $2x = 12$ , so $x = 6$ .

Answer: $x = 6$

Worked Example 3b

Given $f(x) = x^{2} - 2x$ , work out $f(5)$ and $f(0)$ .

For $f(5)$ : $(5)^{2} - 2(5) = 25 - 10 = 15$ . For $f(0)$ : $(0)^{2} - 2(0) = 0$ .

Answer: $f(5) = 15$ and $f(0) = 0$

Evaluating the same function at several inputs is exactly how you would generate points to plot its graph, which links this Higher topic back to the work on graphs.

Composite Functions

A composite function applies two functions in turn. The notation $fg(x)$ means "do $g$ first, then apply $f$ to the result" — work from the inside out. The function written closest to the $x$ always acts first. This ordering catches many students out, so it is worth saying aloud each time: "in $fg$ , $g$ goes first". Think back to the machine picture: $fg(x)$ feeds $x$ into machine $g$ , and whatever comes out is fed into machine $f$ . To build the algebraic expression for $fg(x)$ , take the whole output of $g$ and substitute it wherever $x$ appears in $f$ .

Worked Example 4

Given $f(x) = x + 5$ and $g(x) = 2x$ , work out $fg(3)$ .

Do $g$ first: $g(3) = 2(3) = 6$ . Then apply $f$ : $f(6) = 6 + 5 = 11$ .

Answer: $11$

Worked Example 5

With the same $f$ and $g$ , find an expression for $fg(x)$ .

Do $g$ first: $g(x) = 2x$ . Then $f(2x) = 2x + 5$ .

Answer: $fg(x) = 2x + 5$

Worked Example 6 — order matters

With $f(x) = x + 5$ and $g(x) = 2x$ , find $gf(x)$ and compare with $fg(x)$ .

In $gf(x)$ the inner function is $f$ , so do $f$ first: $f(x) = x + 5$ . Then apply $g$ to that result: $g(x + 5) = 2(x + 5) = 2x + 10$ .

Answer: $gf(x) = 2x + 10$ , which is not the same as $fg(x) = 2x + 5$ — order matters.

This comparison is worth remembering: for most pairs of functions, $fg(x)$ and $gf(x)$ give different expressions. Composition is not like multiplication, where the order makes no difference. Whenever an exam question gives you a composite, the first thing to settle is which function is on the inside.

Common error: applying the functions in the wrong order. In $fg(x)$ the function nearest the $x$ , namely $g$ , acts first.

Inverse Functions

The inverse $f^{-1}(x)$ reverses $f$ : if $f$ turns $3$ into $8$ , then $f^{-1}$ turns $8$ back into $3$ . To find it, write $y = f(x)$ , swap the roles of $x$ and $y$ , then rearrange to make $y$ the subject. The swap is the heart of the method, because finding an inverse is precisely the act of exchanging input and output. A quick way to check your inverse is correct is to confirm that $f(f^{-1}(x)) = x$ — applying a function and then its inverse should return the original input.

Worked Example 7

Find the inverse of $f(x) = 3x - 4$ .

Write $y = 3x - 4$ . Swap: $x = 3y - 4$ . Rearrange: $x + 4 = 3y$ , so $y = \dfrac{x + 4}{3}$ .

Answer: $f^{-1}(x) = \dfrac{x + 4}{3}$

Worked Example 8

Find the inverse of $g(x) = \dfrac{x}{2} + 1$ .

Write $y = \dfrac{x}{2} + 1$ . Swap: $x = \dfrac{y}{2} + 1$ . Subtract $1$ : $x - 1 = \dfrac{y}{2}$ . Multiply by $2$ : $y = 2(x - 1)$ , i.e. $y = 2x - 2$ .

Answer: $g^{-1}(x) = 2x - 2$

Simplifying Algebraic Fractions

An algebraic fraction is simplified by factorising the top and bottom and cancelling any common factor. The golden rule is that you may only cancel a factor — something that multiplies the whole numerator and the whole denominator — never an individual term that is added or subtracted. This is exactly why factorising first is essential: it turns sums into products, exposing the factors that are safe to cancel.

Worked Example 9

Simplify $\dfrac{6x^{2}}{9x}$ .

The HCF of $6$ and $9$ is $3$ , and $x^{2} \div x = x$ : $\dfrac{6x^{2}}{9x} = \dfrac{2x}{3}$ .

Answer: $\dfrac{2x}{3}$

Worked Example 10

Simplify $\dfrac{x^{2} - 9}{x^{2} + 7x + 12}$ .

Factorise both: top is $(x + 3)(x - 3)$ ; bottom is $(x + 3)(x + 4)$ . Cancel the common factor $(x + 3)$ : $\dfrac{x - 3}{x + 4}$ .

Answer: $\dfrac{x - 3}{x + 4}$

Common error: cancelling individual terms rather than whole factors — you may only cancel a factor common to the entire numerator and denominator.

Functions and Algebraic Fractions (Higher)

Functions and Algebraic Fractions (Higher)

Key Vocabulary

Function Notation and Evaluating

Worked Example 1

Worked Example 2

Worked Example 3 — solving with function notation

Worked Example 3b

Composite Functions

Worked Example 4

Worked Example 5

Worked Example 6 — order matters

Inverse Functions

Worked Example 7

Worked Example 8

Simplifying Algebraic Fractions

Worked Example 9

Worked Example 10

Adding and Subtracting Algebraic Fractions

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