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Composition of Functions: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 Two lookup records

Let $f$ be the pairs $\{(0,2),(2,1),(3,0)\}$ and let $g$ be the pairs $\{(0,4),(1,5)\}$ . Evaluate $(g\circ f)(2)$ .
1. Hint 1
  The rightmost function acts on the original input.
2. Hint 2
  Find the output paired with $2$ in $f$ , then use it as the input to $g$ .
Answer

$(g\circ f)(2)=5$ .

Full solution

The record for $f$ gives $f(2)=1$ .
The record for $g$ then gives $g(1)=5$ .
$g(f(2))=5$

The intermediate value $1$ is in the domain of $g$ , so the composition is defined.

Answer

$(g\circ f)(2)=5$ .

Key idea

Composition follows an output from one rule into the input of the next.
Problem 1: Mark complete
Problem 2 A magnitude pipeline

Let $f(x)=2-\lvert x\rvert$ and $g(x)=3x+4$ , both with domain $\mathbb R$ . Write $(g\circ f)(x)$ in simplified form.
1. Hint 1
  The whole output of $f$ becomes the argument of $g$ .
2. Hint 2
  Place $2-\lvert x\rvert$ in the input slot of $3x+4$ .
Answer

$(g\circ f)(x)=10-3\lvert x\rvert$ on $\mathbb R$ .

Full solution

Substitution gives
$\begin{aligned}(g\circ f)(x)&=3(2-\lvert x\rvert)+4\\&=10-3\lvert x\rvert\end{aligned}$

Both rules accept the needed real values, so the domain remains all real numbers.

Answer

$(g\circ f)(x)=10-3\lvert x\rvert$ on $\mathbb R$ .

Key idea

Composition substitutes a full function value into every input slot of another rule.
Problem 2: Mark complete
Problem 3 A three-position cycle

The function $f$ maps $0$ to $2$ , $2$ to $5$ , and $5$ to $0$ , with domain and codomain $\{0,2,5\}$ . Find $f^2(5)$ , where $f^2=f\circ f$ .
1. Hint 1
  The superscript counts applications of the rule.
2. Hint 2
  Follow the arrow from $5$ , then follow the arrow from its output.
Answer

$f^2(5)=2$ .

Full solution

The first application gives $f(5)=0$ .
The second gives $f(0)=2$ .
$\begin{aligned}f^2(5)&=f(0)\\&=2\end{aligned}$

Every intermediate value is in the stated domain.

Answer

$f^2(5)=2$ .

Key idea

An iterate feeds the first output back into the same function.
Problem 3: Mark complete
Problem 4 Two discounts

For any nonnegative price $q$ dollars, define $A(q)=0.8q$ and $B(q)=q-6$ . A starting price is $p\ge10$ dollars. One shop applies $A$ and then $B$ ; another applies $B$ and then $A$ . Write both final-price rules and determine which is lower and by how much.
1. Hint 1
  The operation performed second acts on the entire result of the first.
2. Hint 2
  Compare $B(A(p))$ with $A(B(p))$ before substituting a particular price.
Answer

$B(A(p))=0.8p-6$ ; $A(B(p))=0.8p-4.8$ ; the first shop is lower by $1.20$ dollars.

Full solution

The first shop gives
$B(A(p))=0.8p-6.$
The second gives
$\begin{aligned}A(B(p))&=0.8(p-6)\\&=0.8p-4.8\end{aligned}$

The second price minus the first is $1.2$ dollars.
For $p\ge10$ all prices involved are nonnegative, so both discount orders are meaningful.

Answer

$B(A(p))=0.8p-6$ ; $A(B(p))=0.8p-4.8$ ; the first shop is lower by $1.20$ dollars.

Key idea

Changing composition order can change a practical outcome by a fixed amount.
Problem 4: Mark complete
Problem 5 A bounded pipeline

Let $f(x)=\lvert x\rvert-2$ on $\mathbb R$ and $g(u)=\sqrt{5-u}$ on its natural domain. Find the formula and natural domain of $g\circ f$ .
1. Hint 1
  The inside output must meet the input condition of the square root rule.
2. Hint 2
  Substitute $\lvert x\rvert-2$ for $u$ and require the resulting radicand to be nonnegative.
Answer

$(g\circ f)(x)=\sqrt{7-\lvert x\rvert}$ ; domain $[-7,7]$ .

Full solution

The composed formula is
$g(f(x))=\sqrt{7-\lvert x\rvert}.$

The inner rule accepts every real $x$ .
The outer root requires $7-\lvert x\rvert\ge0$ , so $\lvert x\rvert\le7$ and the domain is $[-7,7]$ .
Both endpoints yield zero and are allowed.

Answer

$(g\circ f)(x)=\sqrt{7-\lvert x\rvert}$ ; domain $[-7,7]$ .

Key idea

A composite domain consists of the inner inputs whose outputs satisfy the outer rule.
Problem 5: Mark complete
Problem 6 Two possible splits

Let $H(x)=\lvert3x+2\rvert-5$ on $\mathbb R$ . Write $H=g\circ f$ in two ways: first with $f(x)=3x+2$ , then with $f(x)=3x$ . State the outer rule and its natural domain in each case.
1. Hint 1
  The outer rule must finish exactly the operations the chosen inner rule has not done.
2. Hint 2
  Recompose your chosen outer rule with the inner rule and match the result term by term against $H(x)$ .
Answer

First: $g(u)=\lvert u\rvert-5$ . Second: $g(u)=\lvert u+2\rvert-5$ . Both outer domains are $\mathbb R$ .

Full solution

With the first inner rule, choose $g(u)=\lvert u\rvert-5$ , yielding
$g(3x+2)=H(x).$
With the second, choose $g(u)=\lvert u+2\rvert-5$ , yielding
$g(3x)=H(x).$

Each outer rule uses absolute value and real arithmetic only, so each has natural domain $\mathbb R$ .
Recomposition verifies both different splits.

Answer

First: $g(u)=\lvert u\rvert-5$ . Second: $g(u)=\lvert u+2\rvert-5$ . Both outer domains are $\mathbb R$ .

Key idea

Decomposition depends on which part of a rule is assigned to the inner function.
Problem 6: Mark complete
Problem 7 A simplified chain

Let $f(x)=\frac{x-1}{x+1}$ and $g(u)=\frac{u+1}{u-1}$ on their natural real domains. Find a simplified formula and the exact domain of $g\circ f$ .
1. Hint 1
  Record the inner exclusion before simplifying the nested fractions.
2. Hint 2
  Then decide whether an output of $f$ can equal the forbidden outer input $1$ .
Answer

$(g\circ f)(x)=-x$ for $x\ne-1$ .

Full solution

The inner function requires $x\ne-1$ .
Its output cannot equal $1$ , since that would require $x-1=x+1$ .
Thus the outer function creates no extra exclusion.

For an allowed input, combine the outer numerator and denominator over the common denominator $x+1$ :
$\begin{aligned}\frac{x-1}{x+1}+1&=\frac{(x-1)+(x+1)}{x+1}\\&=\frac{2x}{x+1}\end{aligned}$ $\begin{aligned}\frac{x-1}{x+1}-1&=\frac{(x-1)-(x+1)}{x+1}\\&=-\frac2{x+1}\end{aligned}$
The second expression is nonzero.
Their quotient is
$g(f(x))=-x.$
The original exclusion $x\ne-1$ remains even though the formula simplifies.

Answer

$(g\circ f)(x)=-x$ for $x\ne-1$ .

Key idea

Simplification of a composite does not erase the domain restrictions of its intermediate steps.
Problem 7: Mark complete
Problem 8 Two composition orders

Let $f(x)=\lvert x\rvert$ and $g(x)=x^2$ , both on $\mathbb R$ with codomain $[0,\infty)$ . A student says these functions commute. Is that correct? Check formulas and domains.
1. Hint 1
  Commuting means both orders produce the same function.
2. Hint 2
  Compare taking a magnitude before squaring with taking a magnitude after squaring.
Answer

Yes; both composites equal $x^2$ on $\mathbb R$ , with codomain $[0,\infty)$ .

Full solution

One order gives
$\begin{aligned}g(f(x))&=\lvert x\rvert^2\\&=x^2\end{aligned}$
The other gives
$\begin{aligned}f(g(x))&=\lvert x^2\rvert\\&=x^2\end{aligned}$

Both inner outputs are accepted by the other function, so each composite has domain $\mathbb R$ .
Their codomains also match.
Therefore the composites are equal.

Answer

Yes; both composites equal $x^2$ on $\mathbb R$ , with codomain $[0,\infty)$ .

Key idea

Some different functions commute, but the formulas and domains must be checked for the particular pair.
Problem 8: Mark complete
Problem 9 A constant function twice

Let $f(x)=2$ for every real $x$ . A student writes $f^2(x)=4$ . Here $f^2$ means $f\circ f$ . Identify the error and give both $f^2(x)$ and $(f(x))^2$ .
1. Hint 1
  Applying a rule twice differs from multiplying its output by itself.
2. Hint 2
  Ask what input $f$ actually receives on its second application, not what the first output looks like.
Answer

$f^2(x)=2$ ; $(f(x))^2=4$ ; the error is squaring the output $2$ instead of applying $f$ a second time.

Full solution

The iterate sends any $x$ to $2$ , then applies $f$ again.
$\begin{aligned}f^2(x)&=f(2)\\&=2\end{aligned}$

The squared output instead multiplies $2$ by itself, giving $4$ .
The student used the meaning of a numerical exponent in place of the stated composition convention.

Answer

$f^2(x)=2$ ; $(f(x))^2=4$ ; the error is squaring the output $2$ instead of applying $f$ a second time.

Key idea

A superscript denoting composition counts applications rather than multiplication of outputs.
Problem 9: Mark complete
Problem 10 An overlap shortcut

Let $f(x)=x+4$ have declared domain $[-2,2]$ , and let $g(u)=\sqrt u$ have domain $[0,\infty)$ . A student claims the domain of $g\circ f$ is $[0,2]$ , the overlap of these domains. Assess the claim and give the correct domain.
1. Hint 1
  The outer function receives the output of the inner function, not the original input.
2. Hint 2
  Find the complete output interval of $f$ on its declared domain.
Answer

False; the composite domain is $[-2,2]$ .

Full solution

For every allowed $x$ ,
$2\le x+4\le6.$
Thus every output of $f$ is nonnegative and accepted by $g$ .

All inner inputs in $[-2,2]$ survive.
For example, $x=-1$ gives the defined output $\sqrt3$ , although $-1$ is outside the claimed overlap.

Answer

False; the composite domain is $[-2,2]$ .

Key idea

Composite domains filter inner outputs through the outer domain instead of intersecting two input sets.
Problem 10: Mark complete

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