SL 2.5 Work with composite and inverse functions | International Baccalaureate (IB) IB AA SL: Mathematics: Analysis and Approaches SL

International Baccalaureate (IB) IB AA SL: Mathematics: Analysis and Approaches SL

Functions

IB Mathematics: Analysis and Approaches SL — Study topic SL 2.5

A function takes an allowed input and assigns an output. If one function doubles an input and another adds

3

, applying one after the other creates a composite function. An inverse function reverses an input-output relationship when that relationship can be reversed uniquely. In this lesson, you will work with both ideas using algebra, numerical values, graphs, and contexts. The order of composition matters, and a function may need a restricted domain before it has an inverse function.

What you will learn

Evaluate and simplify composite functions, paying attention to order and domain.
Find an inverse function when it exists, and state its domain and range.
Interpret composition and inverse functions using algebra, tables, graphs, and contexts.
Check results by substitution and use graphing technology to confirm reasoning.

1. Prior knowledge: function notation and domains

The notation

f(x)

means the output of function

f

for input

x

; it does not mean

f

multiplied by

x

. A function's domain is the set of inputs it accepts, and its range is the set of outputs it produces. Check these restrictions before combining functions.

For example,

g(x)=\sqrt{x-1}

accepts inputs for which

x-1\geq 0

, so its domain is

x\geq 1

. The rule

h(x)=2x+5

accepts every real input. Such restrictions matter when one function's output becomes another function's input.

Function notation can describe a process. If

f

converts a time in hours into a distance and

g

converts a distance into a cost, then

g(f(t))

gives the cost associated with time

t

. The input and output units help identify which order makes sense.

Check the domain of the function receiving the initial input.
An output from one function must be an allowed input for the next.

2. Composite functions: apply one rule, then another

The composite function

g\circ f

is defined by

(g\circ f)(x)=g(f(x))

. Read this as “

g

after

f

”: apply

f

first, then apply

g

to the result. The function written on the inside acts first. Changing the order can change the result, so

g\circ f

and

f\circ g

need not be equal.

To find a rule for a composite function, substitute the entire expression

f(x)

wherever the input variable appears in

g

. Use brackets to keep the substitution clear, then simplify. You can also evaluate at a particular input by following the two functions in order.

The domain of

g\circ f

includes inputs in the domain of

f

whose outputs are in the domain of

g

. This is a two-part check: the starting input must be allowed for

f

, and the intermediate value

f(x)

must be allowed for

g

. A simplified expression does not automatically remove these restrictions.

A table can show an input passing through

f

to give an intermediate value, which then passes through

g

. On a graph, the first function's output becomes the input used for the second function. In a context, check the units at each stage.

(g\circ f)(x)=g(f(x))

In $(g\circ f)(x)$ , evaluate $f$ first.
The composite domain must meet the restrictions of both functions.
The intermediate output of the first function becomes the input of the second.

3. Inverse functions: reverse an input-output rule

An inverse function reverses a function's action. If

f

sends

a

to

b

, its inverse sends

b

back to

a

. The notation is

f^{-1}

; it does not mean the reciprocal

1/f

. Reversing is possible as a function only when each output in the relevant range comes from exactly one input.

For a one-to-one function, write

y=f(x)

, interchange

x

and

y

, and solve for

y

. The resulting rule is

f^{-1}(x)

. The domain of

f^{-1}

is the range of

f

, and the range of

f^{-1}

is the domain of

f

.

A function that gives the same output for two different inputs cannot be reversed uniquely on its full domain. For example,

x^2

gives

4

for both

2

and

-2

. It has no inverse function on all real inputs; restricting its original domain to

x\geq 0

makes each output correspond to one input.

The graphs of a function and its inverse are reflections across the line

y=x

, because reversing an input-output pair

(a,b)

gives

(b,a)

. A graphing calculator can help check whether every horizontal line meets the original graph at most once. This is a visual check; algebra and any needed domain restriction still matter.

f^{-1}(f(x))=x

Swap input and output, then solve for the new output.
The inverse's domain and range are the original function's range and domain, respectively.
A function must be one-to-one on its stated domain to have an inverse function.

4. Technology and checking

A graphing calculator can display a function, a proposed inverse, and the line

y=x

. Check whether the two function graphs appear reflected across that line, using a suitable viewing window. This visual check does not replace finding the inverse algebraically.

For a composite function, compare a calculator's value at a chosen input with a two-step calculation by hand. For an inverse, substitute the proposed rule into the original in both orders where the domains allow it. These checks can reveal a reversed order, an algebra error, or a domain mismatch.

In a written solution, show the substitutions and solving steps, then state relevant domain restrictions. Give decimal approximations only when needed and state their accuracy. In a context, include units in the interpretation.

Use technology to visualize and verify a reasoned result.
A numerical or graphical check supports, but does not replace, algebraic reasoning.

Worked example

1. Find and evaluate composite functions

Let

f(x)=3x-2

and

g(x)=x^2+1

. Find

(g\circ f)(x)

and

(f\circ g)(x)

, then evaluate

(g\circ f)(2)

.

Apply the inside function first
For $g\circ f$ , substitute the complete expression $f(x)$ into the input of $g$ .
$(g\circ f)(x)=(3x-2)^2+1$
Simplify and reverse the order
Expand the square for the first rule. For $f\circ g$ , put $g(x)$ into the input of $f$ instead.
$(g\circ f)(x)=9x^2-12x+5,\quad (f\circ g)(x)=3x^2+1$
Evaluate using the composite rule
Substitute $x=2$ into $g\circ f$ . Equivalently, $f(2)=4$ and then $g(4)=17$ .
$(g\circ f)(2)=9(2)^2-12(2)+5=17$

Answer:

(g\circ f)(x)=9x^2-12x+5

,

(f\circ g)(x)=3x^2+1

, and

(g\circ f)(2)=17

. Both composites have domain all real numbers.

Check: The two composite rules differ, confirming that order matters. Following the two function steps gives

f(2)=4

and

g(4)=17

, as required.

Worked example

2. Find an inverse with a restricted domain

Let

f(x)=x^2-4

with domain

x\geq 0

. Find

f^{-1}(x)

and state its domain and range.

Write the function with an output variable
Use $y$ for the output and retain the given restriction on the original input.
$y=x^2-4,\quad x\geq0$
Interchange input and output
Reversing the input-output pairs means swapping $x$ and $y$ .
$x=y^2-4$
Solve for the new output
Rearrange to get $y^2=x+4$ . The original restriction $x\geq0$ means the inverse output must be non-negative, so choose the non-negative square root.
$y=\sqrt{x+4}$

Answer:

f^{-1}(x)=\sqrt{x+4}

, with domain

x\geq -4

and range

y\geq 0

.

Check: For original inputs

x\geq0

,

f^{-1}(f(x))=\sqrt{x^2}=x

. The range of the original function is

y\geq-4

, which is the domain of the inverse.

Worked example

3. Find a composite rule and its domain in context

A conversion rule changes a temperature

C

in degrees Celsius to

f(C)=1.8C+32

degrees Fahrenheit. A second rule is

g(F)=\sqrt{F-32}

. Find

(g\circ f)(C)

and its domain.

Substitute the conversion output
The Fahrenheit value $f(C)$ becomes the input to $g$ . Substitute the full expression into the square root.
$(g\circ f)(C)=\sqrt{1.8C+32-32}$
Simplify and impose the domain condition
The expression under a square root must be non-negative. Since $1.8$ is positive, solve the resulting inequality for $C$ .
$(g\circ f)(C)=\sqrt{1.8C},\quad C\geq0$

Answer:

(g\circ f)(C)=\sqrt{1.8C}

for

C\geq0

. The output has the units specified by

g

.

Check: At

C=10

, the conversion gives

50

degrees Fahrenheit, and

g(50)=\sqrt{18}\approx4.24

to three significant figures. This agrees with evaluating the composite rule.

Common mistakes and how to avoid them

Reading

(g\circ f)(x)

as

f(g(x))

.

Correction: The function on the inside acts first, so

(g\circ f)(x)=g(f(x))

.

Assuming

f^{-1}(x)

means

1/f(x)

.

Correction: The superscript

-1

denotes the inverse function; a reciprocal is written

1/f(x)

.

Finding an inverse formula but ignoring whether the original function is one-to-one.

Correction: Check whether an output can come from more than one input. If so, restrict the original domain appropriately or state that there is no inverse function on that domain.

Using the original domain as the inverse's domain.

Correction: The inverse's domain is the original function's range, and its range is the original domain.

Lesson summary

A composite function applies the inside function first: $(g\circ f)(x)=g(f(x))$ .
The domain of a composite must satisfy the restrictions of both stages.
To find an inverse, swap input and output and solve; ensure the original function is one-to-one on its domain.
The graphs of a function and its inverse reflect across $y=x$ , and substitution provides an algebraic check.

Check your understanding

Question 1

If

f(x)=2x+1

and

g(x)=x^2

, what is

(g\circ f)(3)

?

$49$
$37$
$7$
$19$

Show answer and explanation

49

First

f(3)=7

, then

g(7)=49

.

Question 2

For

f(x)=5x-10

, what is

f^{-1}(x)

?

$5x+10$
$\frac{x+10}{5}$
$\frac{x-10}{5}$
$10-5x$

Show answer and explanation

\frac{x+10}{5}

From

y=5x-10

, swap variables and solve:

x=5y-10

, so

y=(x+10)/5

.

Question 3

The function

h(x)=x^2

has domain all real numbers. Which statement is correct?

It has an inverse function because every output is positive.
It has no inverse function on all real numbers because different inputs can give the same output.
Its inverse is $1/x^2$ .
Its inverse has domain all real numbers.

Show answer and explanation

It has no inverse function on all real numbers because different inputs can give the same output.

For instance,

h(2)=h(-2)=4

, so the output does not identify a unique input.

Key terms

Composite function: A function formed by applying one function to the output of another, such as $(g\circ f)(x)=g(f(x))$ .
Inverse function: A function that reverses the input-output pairs of an original function when each relevant output comes from exactly one input.
One-to-one: A function for which different inputs in its domain always produce different outputs.
Domain: The set of inputs for which a function is defined.
Range: The set of outputs a function produces from its domain.

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SL 2.5 · Work with composite and inverse functions

International Baccalaureate (IB) IB AA SL: Mathematics: Analysis and Approaches SL

IB Mathematics: Analysis and Approaches SL — Study topic SL 2.5

What you will learn

1. Prior knowledge: function notation and domains

2. Composite functions: apply one rule, then another

3. Inverse functions: reverse an input-output rule

4. Technology and checking

1. Find and evaluate composite functions

2. Find an inverse with a restricted domain

3. Find a composite rule and its domain in context

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Key terms

Continue through IB AA SL

About this lesson and its review