In fg(x), which function do I apply first?

g. The function written next to x acts first, so fg(x) means f(g(x)). Reading it left to right is the most common error on this topic.

Why does the question restrict the domain before asking for an inverse?

Only a one-one function has an inverse. A quadratic sends two inputs to the same output, so the domain is cut at the vertex to keep one side of the curve.

How do I know if fg exists?

fg exists only when every output of g is an input f accepts, which means the range of g must lie inside the domain of f.

9709 P1 Functions Questions: Composite and Inverse

Functions carry a lot of marks on 9709 Paper 1, and almost every question tests one of four skills: build a composite in the right order, find an inverse, restrict a domain so an inverse exists, or decide whether a composite exists at all. There is one question on each below.

Question 1: a composite equation [4 marks]

$f (x) = 2 x - 3$ and $g (x) = x^{2} + 4 x$ , both for $x \in R$ . Solve $g f (x) = 12$ .

Show the worked solution

In $g f (x)$ the letter next to $x$ is $f$ , so $f$ acts first and $g$ acts on its output.

g f (x) = g (2 x - 3) = (2 x - 3)^{2} + 4 (2 x - 3)

Expand both parts, then collect:

= 4 x^{2} - 12 x + 9 + 8 x - 12 = 4 x^{2} - 4 x - 3

Set it equal to 12 and factorise:

4 x^{2} - 4 x - 15 = 0 ⟹ (2 x - 5) (2 x + 3) = 0 ⟹ x = \frac{5}{2} or x = - \frac{3}{2}

Check $x = 2.5$ : $f (2.5) = 2$ and $g (2) = 4 + 8 = 12$ .

Building $f g$ instead gives $2 x^{2} + 8 x - 3$ , a different equation with different roots. The order is where the marks are.

Question 2: finding an inverse [3 marks]

$h (x) = \frac{2 x + 1}{x - 3}$ for $x \neq = 3$ . Find $h^{- 1} (x)$ .

Show the worked solution

Write $y = h (x)$ and make $x$ the subject. The $x$ appears twice, so collect the $x$ terms on one side and factorise.

y (x - 3) = 2 x + 1 ⟹ x y - 3 y = 2 x + 1 ⟹ x y - 2 x = 3 y + 1

x (y - 2) = 3 y + 1 ⟹ x = \frac{3 y + 1}{y - 2}

Swap the letters:

h^{- 1} (x) = \frac{3 x + 1}{x - 2}, x \neq = 2

Check: $h (4) = \frac{9}{1} = 9$ and $h^{- 1} (9) = \frac{28}{7} = 4$ .

Question 3: restrict the domain, then invert [6 marks]

$f (x) = x^{2} - 6 x + 11$ for $x \geq k$ .

(a) Express $f (x)$ in the form $(x - a)^{2} + b$ .

(b) State the smallest value of $k$ for which $f$ has an inverse.

Show the worked solution

(a) $x^{2} - 6 x + 11 = (x - 3)^{2} - 9 + 11 = (x - 3)^{2} + 2$ .

(b) The vertex is at $x = 3$ . To the right of it the curve only rises, so each output comes from one input. $k = 3$ .

(c) Let $y = (x - 3)^{2} + 2$ . Then $(x - 3)^{2} = y - 2$ , so $x - 3 = \pm y - 2$ .

The domain is $x \geq 3$ , so $x - 3 \geq 0$ and only the positive root applies:

f^{- 1} (x) = 3 + x - 2

The domain of $f^{- 1}$ is the range of $f$ . With $x \geq 3$ , the smallest output is $f (3) = 2$ , so the domain of $f^{- 1}$ is $x \geq 2$ .

Check: $f (5) = 4 + 2 = 6$ and $f^{- 1} (6) = 3 + 4 = 5$ .

Question 4: does fg exist? [4 marks]

$f (x) = x + 3$ for $x \geq - 3$ , and $g (x) = x^{2} + 4 x$ for $x \geq 0$ .

(a) Explain why $f g$ exists.

(b) Find the range of $f g$ .

Show the worked solution

(a) $f g$ exists when every output of $g$ is an input $f$ accepts: the range of $g$ must lie inside the domain of $f$ .

For $x \geq 0$ , $g$ is increasing (its vertex is at $x = - 2$ , to the left of the domain) and $g (0) = 0$ . So the range of $g$ is $g (x) \geq 0$ .

Every value $\geq 0$ is also $\geq - 3$ , so the range of $g$ lies inside the domain of $f$ , and $f g$ exists.

(b) $f g (x) = x^{2} + 4 x + 3$ . The inside is smallest at $x = 0$ , where it is 3, and grows after that. So the range of $f g$ is $f g (x) \geq 3$ .

Without the restriction $x \geq 0$ , $g$ reaches $- 4$ at $x = - 2$ , and $f (- 4)$ is undefined. That is why the question states the domain of $g$ .

Try it: watch an input go through both functions

Move $x$ and see which values of $g (x)$ the square root in $f$ can accept.

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A Level Maths 9709 Paper 1: functions exam questions

Question 1: a composite equation [4 marks]

Question 2: finding an inverse [3 marks]

Question 3: restrict the domain, then invert [6 marks]

Question 4: does fg exist? [4 marks]

Try it: watch an input go through both functions

Questions students ask