Values of f, f∘g and f⁻¹ from a graph or a formula

In 8 of 20 recent exams
Paper 1, no calculator
1 to 3 marks

How it is set in the exam

Paper 1 (no calculator), 1–3 marks per part and up to about 5 in one question (Examples 1.1 and 1.3 combine more parts, for practice): usually the first question of Section A, built on a graph, or the opening parts of a Section B question on functions. Common (8 of 20 exam versions, all on Paper 1 (no calculator)). A value read in one part is often fed into a composite or an inverse in the next.

Spot it in the exam

Exam phrases to spot (5)

“write down the value of $f(3)$ ”
“find $(f \circ g)(0)$ ”
“find $(f \circ f)(4)$ ”
“find $f^{-1}(2)$ ”
“find the value of $x$ for which $f^{-1}(2x - 5) = 1$ ”

In disguise (4)

"find the value of $x$ such that $f(x) = 5$ " for a one-to-one $f$ given by a graph or a table: that $x$ is $f^{-1}(5)$ , read across from $5$
the values of $f$ and $g$ are given in a table instead of a graph
the exact value of $f^{-1}(b)$ for an exponential $f$ , before any formula for $f^{-1}$ has been found: solve $f(x) = b$ with a common base
$f^{-1}(b)$ where an earlier part found $f(a) = b$ : then $f^{-1}(b) = a$ , with nothing to work out

Variants

a composite of a graph and a formula; a composite with an inverse inside, $(f^{-1} \circ g)(a)$ or $f^{-1}(f^{-1}(b))$ (not set 2021–26); the domain or range of $f^{-1}$ read from the graph of $f$ when no sketch of $f^{-1}$ is asked (the table in Finding an inverse routes it); a chain that opens by writing a linear function from given information (that part is Equation of a straight line), then uses it for values.

Looks like it, but isn't (4)

the formula $f^{-1}(x)$ is asked for: Find the inverse function, with its domain
a composite expression is needed, or a constant in one: Form a composite function and use it
the formula for $f^{-1}$ has already been found, written down or given, and is a log, and now a value of it is asked for: Exact value of a logarithm: write it as a power (Topic 1)
$f(x) = b$ solved from a formula, when no inverse is mentioned (in symbols or in words) and there is no composite: the equation's own type, such as Solve an exponential equation (unknown in the power, or in the base), or a model reaching a level (Exponential and log models: initial value, rate constant, value, time)

Worked example

From a graph: a value, f∘f, f⁻¹ and the domain of f⁻¹

Paper 1 (no calculator)

Question

The graph of $y = f(x)$ , for $-2 \le x \le 8$ , consists of two line segments: one from $(-2, -1)$ to $(2, 5)$ , and one from $(2, 5)$ to $(8, 8)$ .
(a) Write down $f(0)$ . [1]
(b) Find $(f \circ f)(0)$ . [2]
(c) Find $f^{-1}(6)$ . [2]
(d) Write down the domain of $f^{-1}$ . [1]

To see the picture: join the three points on squared paper. The exam shows the graph on a grid, and every value here is read off it; the segment equations in the working are only a check.

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Steps

A value of $f$ :
read it off the graph (from $x$ on the horizontal axis, go to the curve, then across to the $y$ -axis), or substitute into the formula.
A composite value, $(f \circ g)(a)$ :
inside first. Find $g(a)$ , then $f$ of that number. Write the middle step, e.g. $f(g(2)) = f(-1)$ : it carries the method mark.
An inverse value, $f^{-1}(b)$ :
find the input that gives the output $b$ . On a graph, go across from $b$ on the $y$ -axis to the curve, then to the $x$ -axis; with a formula, solve $f(x) = b$ . No formula for $f^{-1}$ is needed.
An inverse equation, $f^{-1}(\text{expression}) = c$ :
apply $f$ to both sides, so $\text{expression} = f(c)$ . Then solve.
A composite equal to a number, $(f \circ g)(x) = c$ , with $f$ and $g$ given by a graph or a table:
work backwards, outer function first. Find every input $u$ with $f(u) = c$ (read across from $c$ ), then every $x$ with $g(x) = u$ (Your turn 4(a)). With formulas, form the composite and solve it: Form a composite function and use it.
The domain or range of $f^{-1}$ :
swap those of $f$ . The domain of $f^{-1}$ is the range of $f$ , the $y$ -values its graph takes. For an unbroken graph, that is from its lowest to its highest $y$ -value, with $\le$ at a closed dot, $<$ at an open dot or an asymptote, and no bound at an end where the graph goes on for ever (for $f(x) = 2 \times 9^x$ : above the asymptote $y = 0$ , with no top, so the range is $f(x) > 0$ and the domain of $f^{-1}$ is $x > 0$ ). A graph of $\frac{ax + b}{cx + d}$ on its largest domain takes every value except its horizontal asymptote (Find the inverse function, with its domain). Interval notation, such as $[-1, 8]$ for $-1 \le x \le 8$ , also scores.
Check an inverse value forwards:
if $f^{-1}(4) = 3$ , then $f(3)$ must be $4$ . On , or to check any value, define the function and evaluate it.

Common mistakes

The wrong order in a composite
$(f \circ g)(2)$ is $f(g(2))$ , with $g$ first.
Reading $f^{-1}(b)$ as $f(b)$ , or as $\frac{1}{f(b)}$
$f^{-1}(4)$ is an input of $f$ , found by starting at $4$ on the $y$ -axis.
The domain of $f^{-1}$ from the $x$ -values of $f$
it is the range of $f$ , the $y$ -values.

More mistakes (4)

Dropping the middle step
write $f(g(2)) = f(-1)$ before the value. With a misread graph and no working, nothing scores.
Finding the formula for $f^{-1}$ when a value will do
it scores, but it takes longer and gives more chances to slip.
Applying $f^{-1}$ instead of $f$
in $f^{-1}(2x - 5) = 1$ , $f$ removes the inverse, so $2x - 5 = f(1)$ , not $f^{-1}(1)$ .
The wrong inequality at an end
$\le$ at a closed dot (the end is included), $<$ at an open dot or an asymptote.

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