Composite functions and inverse functions AI HL

IB Mathematics: Applications and Interpretation HL · Unit 2: Functions

Lesson 2.7 · Composite Functions and Inverse Functions

Combining processes, reversing processes and checking domains.

Faith connection

Paul encouraged believers to “Bear ye one another’s burdens” (Galatians 6:2, KJV). Loving someone often involves a sequence of actions: we listen, understand and then help. The order of those actions matters, just as the order matters when we combine functions. As you think about caring for others, come to Jesus, our Savior and Lord, and ask him to teach you patient, practical love.

Learning objectives

By the end of this lesson, you should be able to:

calculate and interpret $(f\circ g)(x)=f(g(x))$ ;
distinguish $(f\circ g)(x)$ from $(g\circ f)(x)$ ;
find the domain of a composite function;
find inverse functions and state necessary domain restrictions;
verify that a function and its inverse undo one another;
solve contextual problems involving several steps and their inverses.

1. Composite functions: one operation after another

A composite function uses the output of one function as the input of another. The notation $(f\circ g)(x)$ means apply $g$ first, then apply $f$ :

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

The function closest to $x$ acts first.

Worked example.

Let $f(x)=3x+2$ and $g(x)=x^2-1$ . Find both composites.

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

$(g\circ f)(x)=g(3x+2)=(3x+2)^2-1=9x^2+12x+3$

The answers differ because changing the order changes the operations.

2. Evaluate a composite at a particular input

Work from the inside out. Calculate the inner output first, then use it as the input for the outer function.

Worked example.

Let $f(x)=\sqrt{x+4}$ and $g(x)=2x-1$ . Find $(f\circ g)(3)$ .

$g(3)=2(3)-1=5$

$(f\circ g)(3)=f(5)=\sqrt{5+4}=3$

Equivalently, $(f\circ g)(x)=\sqrt{2x+3}$ and substituting $x=3$ gives $3$ .

3. Domain of a composite function

To form $f(g(x))$ , two conditions must hold: $x$ must be allowed in $g$ , and the resulting value $g(x)$ must be allowed in $f$ . Checking only the final-looking formula can hide a restriction from the original functions.

Worked example.

Let $f(x)=\sqrt{x-2}$ , with domain $x\geq2$ , and $g(x)=5-2x$ , with domain $\mathbb R$ . Find the domain of $f\circ g$ .

$(f\circ g)(x)=\sqrt{(5-2x)-2}=\sqrt{3-2x}$

The input to $f$ must be at least $2$ :

$5-2x\geq2\quad\Longrightarrow\quad x\leq\frac32$

Therefore the domain of $f\circ g$ is $x\leq\frac32$ .

Worked example · A cancelled factor.

Let $f(x)=1/x$ , with $x\neq0$ , and $g(x)=x-4$ . Then:

$(f\circ g)(x)=\frac{1}{x-4},\qquad x\neq4$

The input $x=4$ is excluded because $g(4)=0$ , which cannot be used as an input to $f$ .

4. Composite functions in context

A chain of real-world processes can be represented by a composite. State what each function accepts and produces so the order is clear.

Worked example · Discount, then tax.

Let $D(p)=0.80p$ give the price in euros after a $20\%$ discount. Let $T(p)=1.10p$ add a $10\%$ tax to the price it receives. Applying the discount first gives:

$(T\circ D)(p)=T(0.80p)=1.10(0.80p)=0.88p$

For an original price of €50, the final price is $0.88(50)=€44$ . In this particular example, reversing the two percentage multiplications produces the same result. That does not mean composites always commute.

Worked example · Order changes the result.

A delivery company adds a fixed €5 fee, represented by $F(p)=p+5$ . A voucher then removes $20\%$ of the amount presented, represented by $D(p)=0.8p$ . For a €40 purchase:

$(D\circ F)(40)=0.8(40+5)=€36$

$(F\circ D)(40)=0.8(40)+5=€37$

A discount applied to the fee as well gives a different price from a discount applied only to the purchase.

5. What is an inverse function?

An inverse function reverses a function. If $f$ takes $a$ to $b$ , then $f^{-1}$ takes $b$ back to $a$ . The notation $f^{-1}(x)$ means the inverse of $f$ ; it does not mean $1/f(x)$ .

When both compositions are defined:

$(f\circ f^{-1})(x)=x,\qquad(f^{-1}\circ f)(x)=x$

The domain of $f^{-1}$ is the range of $f$ , and the range of $f^{-1}$ is the domain of $f$ . Graphs of a function and its inverse reflect across the line $y=x$ .

Worked example.

Let $f(x)=4x-7$ . To find its inverse, write $y=4x-7$ and rearrange:

$y=4x-7\quad\Longrightarrow\quad x=\frac{y+7}{4}$

$f^{-1}(x)=\frac{x+7}{4}$

Check by composition:

$f\left(f^{-1}(x)\right)=4\left(\frac{x+7}{4}\right)-7=x$

6. Why a domain restriction may be needed

For an inverse to be a function, each output of the original function must correspond to only one original input. If a graph gives the same output at two inputs, restrict its domain to a part where it is one-to-one.

Worked example · A restricted quadratic.

Let $f(x)=(x-3)^2-2$ . Over all real $x$ , $f(2)=f(4)=-1$ , so there is no inverse function on that domain. Restrict $f$ to $x\geq3$ .

Starting with $y=(x-3)^2-2$ :

$y+2=(x-3)^2$

Since $x\geq3$ , we have $x-3\geq0$ , so take the positive square root:

$x=3+\sqrt{y+2}$

$f^{-1}(x)=3+\sqrt{x+2},\qquad x\geq-2$

The original range is $[-2,\infty)$ , so it becomes the inverse domain.

Worked example · The other branch.

If instead $f(x)=(x-3)^2-2$ is restricted to $x\leq3$ , then $x-3\leq0$ , so:

$f^{-1}(x)=3-\sqrt{x+2},\qquad x\geq-2$

The chosen restriction determines which square-root branch is correct.

7. Finding the inverse of a more complicated function

State the original domain first. Rearrange carefully, then exchange the input and output roles.

Worked example · A rational function.

Let $f(x)=\dfrac{2x+3}{x-1}$ , where $x\neq1$ . Set $y=f(x)$ :

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

$xy-y=2x+3\quad\Longrightarrow\quad x(y-2)=y+3$

$x=\frac{y+3}{y-2}$

Therefore:

$f^{-1}(x)=\frac{x+3}{x-2},\qquad x\neq2$

The inverse excludes $2$ because $f(x)$ can never equal $2$ : solving $(2x+3)/(x-1)=2$ would require $3=-2$ .

8. Inverses of composites

When both inverses exist on the relevant domains, reversing a two-stage process requires undoing the last step first:

$(f\circ g)^{-1}=g^{-1}\circ f^{-1}$

Worked example.

Let $g(x)=x+2$ and $f(x)=3x$ , both on $\mathbb R$ . Then $(f\circ g)(x)=3(x+2)=3x+6$ .

To undo the result, first divide by $3$ , then subtract $2$ :

$(f\circ g)^{-1}(x)=\frac{x}{3}-2$

Indeed, $f^{-1}(x)=x/3$ and $g^{-1}(x)=x-2$ , so:

$(g^{-1}\circ f^{-1})(x)=g^{-1}\left(\frac{x}{3}\right)=\frac{x}{3}-2$

Calculator and exam tips

For $f(g(x))$ , calculate the inside function first.
Check both domains when composing functions.
To test whether an inverse is possible, inspect whether a horizontal line can cross the relevant part of the original graph more than once.
When reversing a quadratic, use the stated domain to choose the correct square-root sign.
Verify inverses by substitution into the original function.
When using a graphing calculator to solve $f(g(x))=k$ , graph $y=f(g(x))$ and $y=k$ , then find their intersections within the permitted domain.

IB-style Paper 1 · Skill practice

Question 1 · Two orders

Let $f(x)=x^2+2$ and $g(x)=3x-1$ .

Find and simplify $(f\circ g)(x)$ .
Find and simplify $(g\circ f)(x)$ .
Evaluate $(f\circ g)(2)$ and $(g\circ f)(2)$ .

Show solution

$(f\circ g)(x)=(3x-1)^2+2=9x^2-6x+3$ . Also, $(g\circ f)(x)=3(x^2+2)-1=3x^2+5$ . At $x=2$ , the values are $27$ and $17$ , respectively.

Question 2 · Composite domain

Let $f(x)=\sqrt{x+1}$ and $g(x)=\dfrac{6}{x-2}$ . Find $(f\circ g)(x)$ and state its real domain.

Show solution

$(f\circ g)(x)=\sqrt{\dfrac{6}{x-2}+1}=\sqrt{\dfrac{x+4}{x-2}}$ . We require $x\neq2$ and $(x+4)/(x-2)\geq0$ . A sign chart with critical values $-4$ and $2$ gives $x\leq-4$ or $x>2$ . At $x=-4$ , the square root is zero, so that endpoint is included.

Question 3 · Inverse and restriction

The function $h(x)=(x+4)^2+1$ has domain $x\geq-4$ . Find its range and $h^{-1}(x)$ , including the inverse domain.

Show solution

The minimum of $h$ is $1$ , so its range is $h(x)\geq1$ . From $y=(x+4)^2+1$ and $x\geq-4$ , we obtain $x=-4+\sqrt{y-1}$ . Hence $h^{-1}(x)=-4+\sqrt{x-1}$ for $x\geq1$ .

Question 4 · Verify an inverse

Let $p(x)=\dfrac{5x-2}{3}$ and $q(x)=\dfrac{3x+2}{5}$ . Show by composition that $q=p^{-1}$ .

Show solution

$p(q(x))=\frac{5\left(\frac{3x+2}{5}\right)-2}{3}=x$

$q(p(x))=\frac{3\left(\frac{5x-2}{3}\right)+2}{5}=x$

Both functions have domain and range $\mathbb R$ , so the identities establish that $q=p^{-1}$ .

Question 5 · Solve using an inverse

Let $f(x)=\sqrt{2x-5}$ for $x\geq2.5$ .

Find $f^{-1}$ and its domain.
Use your answer to solve $f(x)=7$ .

Show solution

From $y=\sqrt{2x-5}$ , square and rearrange: $y^2=2x-5$ , so $x=(y^2+5)/2$ . Therefore $f^{-1}(x)=(x^2+5)/2$ for $x\geq0$ . Applying it to $7$ gives $x=f^{-1}(7)=(49+5)/2=27$ .

Question 6 · Inverse of a composite

Let $g(x)=2x-5$ and $f(x)=x^3$ , both with domain $\mathbb R$ . Find $(f\circ g)(x)$ and its inverse.

Show solution

$(f\circ g)(x)=(2x-5)^3$ . Set $y=(2x-5)^3$ ; then $\sqrt[3]{y}=2x-5$ , so $x=(\sqrt[3]{y}+5)/2$ . Therefore $(f\circ g)^{-1}(x)=(\sqrt[3]{x}+5)/2$ . Both functions have domain and range $\mathbb R$ .

Applications and interpretation

Application 1 · Conversion and a service charge

For this question, an illustrative conversion function $E(p)=0.90p$ converts an amount $p$ in US dollars to euros. A service adds a €4 charge after conversion, represented by $S(e)=e+4$ .

Find $(S\circ E)(p)$ and explain its meaning.
Find the total euro cost for USS\circ E(S\circ E)(p)=0.90p+480, the total is .

$(S\circ E)^{-1}(x)=\frac{x-4}{0.90}$

For €112, the original amount is $(112-4)/0.90=120$ US dollars. The exchange rate is supplied solely for this exercise.

Application 2 · A discounted ticket and booking fee

A ticket has list price $p$ euros. A student discount reduces it by $15\%$ , represented by $D(p)=0.85p$ . A €3 booking fee is then added, represented by $B(p)=p+3$ .
1. Write a function for the final amount paid.
2. Find the final amount for a €48 ticket.
3. If a student paid €54, find the original list price.
4. Explain why $(B\circ D)(p)$ and $(D\circ B)(p)$ differ.
Show solution

The final amount is $(B\circ D)(p)=0.85p+3$ . For a €48 ticket, the student pays €43.80. If the payment is €54, solve $0.85p+3=54$ , giving $p=60$ euros. Reversing the order gives $(D\circ B)(p)=0.85(p+3)$ , which also discounts the booking fee; the original process does not.

Application 3 · Recovering time from height

A ball’s height is modelled by $H(t)=-(t-2)^2+10$ metres. Consider only the rising part of its flight, $0\leq t\leq2$ seconds.
1. Find the range of $H$ on the stated domain.
2. Find $H^{-1}(x)$ with its domain.
3. When does the ball first reach $9$ m?
Show solution

On $0\leq t\leq2$ , the height increases from $H(0)=6$ m to $H(2)=10$ m, so the range is $6\leq H\leq10$ . From $y=-(t-2)^2+10$ , obtain $(t-2)^2=10-y$ . As $t\leq2$ , choose $t-2=-\sqrt{10-y}$ :

$H^{-1}(x)=2-\sqrt{10-x},\qquad 6\leq x\leq10$

Thus the ball first reaches $9$ m at $H^{-1}(9)=1$ second.

Application 4 · A two-stage manufacturing process

A machine cuts a square sheet with side length $x$ cm into a smaller square by removing $2$ cm from each edge. Its new side length is $L(x)=x-4$ , where $x>4$ . The area function is $A(s)=s^2$ for $s>0$ .
1. Find the final area as a function of the original side length.
2. State its domain and range.
3. Find its inverse and interpret what it calculates.
4. Find the original side length when the final area is $169$ cm $^2$ .
Show solution

$(A\circ L)(x)=(x-4)^2$ for $x>4$ , with range $y>0$ . Because $x-4>0$ , the inverse uses the positive square root:

$(A\circ L)^{-1}(y)=4+\sqrt{y},\qquad y>0$

It gives the required original side length from the final square’s area. For $169$ cm $^2$ , the original side length is $4+\sqrt{169}=17$ cm.

Paper 1 checklist

Read compositions from the inside out. State which operation happens first in a context. Check the domain of every stage. Restrict a function’s domain before assigning it an inverse when necessary. For a proposed inverse, check that both compositions return the original input on their appropriate domains.

Lesson 2.7 · Composite Functions and Inverse Functions

Faith connection

Learning objectives

1. Composite functions: one operation after another

2. Evaluate a composite at a particular input

3. Domain of a composite function

4. Composite functions in context

5. What is an inverse function?

6. Why a domain restriction may be needed

7. Finding the inverse of a more complicated function

8. Inverses of composites

Calculator and exam tips

IB-style Paper 1 · Skill practice

Question 1 · Two orders

Question 2 · Composite domain

Question 3 · Inverse and restriction

Question 4 · Verify an inverse

Question 5 · Solve using an inverse

Question 6 · Inverse of a composite

Applications and interpretation

Application 1 · Conversion and a service charge

Application 2 · A discounted ticket and booking fee

Application 3 · Recovering time from height

Application 4 · A two-stage manufacturing process

Paper 1 checklist

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IB Mathematics: Applications and Interpretation HL 6 Sections80 LessonsLifetime

Curriculum

Composite functions and inverse functions AI HL

Faith connection

Learning objectives

1. Composite functions: one operation after another

2. Evaluate a composite at a particular input

3. Domain of a composite function

4. Composite functions in context

5. What is an inverse function?

6. Why a domain restriction may be needed

7. Finding the inverse of a more complicated function

8. Inverses of composites

Calculator and exam tips

IB-style Paper 1 · Skill practice

Question 1 · Two orders

Question 2 · Composite domain

Question 3 · Inverse and restriction

Question 4 · Verify an inverse

Question 5 · Solve using an inverse

Question 6 · Inverse of a composite

Applications and interpretation

Application 1 · Conversion and a service charge

Application 2 · A discounted ticket and booking fee

Application 3 · Recovering time from height

Application 4 · A two-stage manufacturing process

Paper 1 checklist

Related

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alphy school publishers

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