Pure Mathematics 1

Mathematics9709

Pure Mathematics 1

Mathematics9709

Composite Functions

A joins two rules: the first output becomes the next input. The notation $\mathrm{gf}$ does not mean $\mathrm{g}\times\mathrm{f}$ .

$\mathrm{gf}(x)$ means $\mathrm{g}(\mathrm{f}(x))$ . The inner (right-most) function acts first. Keep its whole output in brackets when you substitute it into the outer rule.

The right-most function acts first

In $\mathrm{gf}(x)$ , the function nearest the $x$ goes first. So $\mathrm{gf}(x)$ says: put $x$ through $\mathrm{f}$ , then put that result through $\mathrm{g}$ . Read it from the inside out.

Worked example

First use numbers: $\mathrm{f}(x)=x+1$ and $\mathrm{g}(x)=3x$

For $\mathrm{gf}(2)$ , do $\mathrm{f}$ first: $2\to3$ , then do $\mathrm{g}$ : $3\to9$ . So $\mathrm{gf}(2)=9$ .

In the other order, $\mathrm{fg}(2)$ gives $2\to6\to7$ . The same input gives a different output because the order changed.

Answer

\mathrm{gf}(2)=9,\qquad \mathrm{fg}(2)=7

In general $\mathrm{gf}(x)\ne\mathrm{fg}(x)$ . Never swap the order unless you have checked that the two results are equal.

How to build one

Worked example

Textbook 2.4: $\mathrm{f}(x)=2x+3$ and $\mathrm{g}(x)=x^2-1$ , find $\mathrm{fg}(x)$ and $\mathrm{gf}(x)$

$\mathrm{fg}(x)$ : do $\mathrm{g}$ first, so put $x^2-1$ into the “double and add 3” machine: $\mathrm{fg}(x)=2(x^2-1)+3$ .
Expand and tidy: $2x^2-2+3=2x^2+1$ .
$\mathrm{gf}(x)$ : do $\mathrm{f}$ first, so put $2x+3$ into the “square and subtract 1” machine, keeping it in a bracket: $\mathrm{gf}(x)=(2x+3)^2-1$ .
Expand the square first, then subtract: $(4x^2+12x+9)-1=4x^2+12x+8$ .

Answer

\mathrm{fg}(x)=2x^2+1

\mathrm{gf}(x)=4x^2+12x+8

If asked to solve $\mathrm{gf}(x)=8$ , form the composite first, then solve the resulting equation: $4x^2+12x+8=8\Rightarrow4x(x+3)=0$ , so $x=0$ or $x=-3$ .

If the question gives a number, evaluate one stage at a time. For the function $\mathrm{f}(x)=\dfrac{x+3}{x-1}$ , write $\mathrm{f}(5)=2$ and then $\mathrm{f}(2)=5$ , so $\mathrm{ff}(5)=5$ . The middle value proves that f was used twice. (9709/12 Oct/Nov 2021 Q3(a))

Find the smallest k so the composite can be formed

A composite exists over the whole stated domain only when every output of the inner function is an allowed input of the outer function: inner range inside outer domain.

Worked example

$\mathrm{f}(x)=2x-3$ for $x\ge k$ , $\mathrm{g}(x)=x^2-4$ for $x\ge-4$ : find the smallest $k$ for which $\mathrm{gf}$ can be formed

$\mathrm{f}(x)=2x-3$ is an increasing line, so on $x\ge k$ its smallest output is at $x=k$ : $\mathrm{f}(k)=2k-3$ .
Force that lowest output into $\mathrm{g}$ 's domain $x\ge-4$ : $2k-3\ge -4$ .
Solve: $2k\ge -1\;\Rightarrow\; k\ge -\tfrac12$ , so the smallest allowed value is the boundary.

Answer

k=-\tfrac12

(9709-style)

Two checks before you finish

Common mistake

Read from the inside: in

\mathrm{gf}(x)

, f acts first. When substituting an expression, enclose the whole expression in brackets before expanding. (9709/12 May/June 2020 Q5)

Common mistake

Before forming the rule, check inner range against outer domain. This condition can be the whole point of a smallest-

k

question.

Now you try

Your turn9709/12 May/June 2020 Q5(a)

The function $\mathrm{f}(x)=a-2x$ is defined for $x\in\mathbb{R}$ , where $a$ is a constant. Express $\mathrm{ff}(x)$ in terms of $a$ and $x$ . (9709/12 May/June 2020 Q5)

Show worked answer

Do $\mathrm{f}$ twice. Put the inner $\mathrm{f}(x)=a-2x$ in for the input, inside a bracket:

$\mathrm{ff}(x)=a-2(a-2x)$

Now expand the bracket and simplify:

$=a-2a+4x$

Answer

\mathrm{ff}(x)=4x-a

PreviousThe Range of a Function NextInverse Functions

Composite Functions

A joins two rules: the first output becomes the next input. The notation $\mathrm{gf}$ does not mean $\mathrm{g}\times\mathrm{f}$ .

$\mathrm{gf}(x)$ means $\mathrm{g}(\mathrm{f}(x))$ . The inner (right-most) function acts first. Keep its whole output in brackets when you substitute it into the outer rule.

The right-most function acts first

In $\mathrm{gf}(x)$ , the function nearest the $x$ goes first. So $\mathrm{gf}(x)$ says: put $x$ through $\mathrm{f}$ , then put that result through $\mathrm{g}$ . Read it from the inside out.

Worked example

First use numbers: $\mathrm{f}(x)=x+1$ and $\mathrm{g}(x)=3x$

For $\mathrm{gf}(2)$ , do $\mathrm{f}$ first: $2\to3$ , then do $\mathrm{g}$ : $3\to9$ . So $\mathrm{gf}(2)=9$ .

In the other order, $\mathrm{fg}(2)$ gives $2\to6\to7$ . The same input gives a different output because the order changed.

Answer

\mathrm{gf}(2)=9,\qquad \mathrm{fg}(2)=7

In general $\mathrm{gf}(x)\ne\mathrm{fg}(x)$ . Never swap the order unless you have checked that the two results are equal.

How to build one

Worked example

Textbook 2.4: $\mathrm{f}(x)=2x+3$ and $\mathrm{g}(x)=x^2-1$ , find $\mathrm{fg}(x)$ and $\mathrm{gf}(x)$

$\mathrm{fg}(x)$ : do $\mathrm{g}$ first, so put $x^2-1$ into the “double and add 3” machine: $\mathrm{fg}(x)=2(x^2-1)+3$ .
Expand and tidy: $2x^2-2+3=2x^2+1$ .
$\mathrm{gf}(x)$ : do $\mathrm{f}$ first, so put $2x+3$ into the “square and subtract 1” machine, keeping it in a bracket: $\mathrm{gf}(x)=(2x+3)^2-1$ .
Expand the square first, then subtract: $(4x^2+12x+9)-1=4x^2+12x+8$ .

Answer

\mathrm{fg}(x)=2x^2+1

\mathrm{gf}(x)=4x^2+12x+8

If asked to solve $\mathrm{gf}(x)=8$ , form the composite first, then solve the resulting equation: $4x^2+12x+8=8\Rightarrow4x(x+3)=0$ , so $x=0$ or $x=-3$ .

Find the smallest k so the composite can be formed

A composite exists over the whole stated domain only when every output of the inner function is an allowed input of the outer function: inner range inside outer domain.

Worked example

$\mathrm{f}(x)=2x-3$ for $x\ge k$ , $\mathrm{g}(x)=x^2-4$ for $x\ge-4$ : find the smallest $k$ for which $\mathrm{gf}$ can be formed

$\mathrm{f}(x)=2x-3$ is an increasing line, so on $x\ge k$ its smallest output is at $x=k$ : $\mathrm{f}(k)=2k-3$ .
Force that lowest output into $\mathrm{g}$ 's domain $x\ge-4$ : $2k-3\ge -4$ .
Solve: $2k\ge -1\;\Rightarrow\; k\ge -\tfrac12$ , so the smallest allowed value is the boundary.

Answer

k=-\tfrac12

(9709-style)

Two checks before you finish

Common mistake

Read from the inside: in

\mathrm{gf}(x)

, f acts first. When substituting an expression, enclose the whole expression in brackets before expanding. (9709/12 May/June 2020 Q5)

Common mistake

Before forming the rule, check inner range against outer domain. This condition can be the whole point of a smallest-

k

question.

Now you try

Your turn9709/12 May/June 2020 Q5(a)

The function $\mathrm{f}(x)=a-2x$ is defined for $x\in\mathbb{R}$ , where $a$ is a constant. Express $\mathrm{ff}(x)$ in terms of $a$ and $x$ . (9709/12 May/June 2020 Q5)

Show worked answer

Do $\mathrm{f}$ twice. Put the inner $\mathrm{f}(x)=a-2x$ in for the input, inside a bracket:

$\mathrm{ff}(x)=a-2(a-2x)$

Now expand the bracket and simplify:

$=a-2a+4x$

Answer

\mathrm{ff}(x)=4x-a

PreviousThe Range of a Function NextInverse Functions