Core practice ← Back to lesson

Combining and Composing 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: Advanced (beyond the core course) Advanced. This problem set goes beyond core Algebra I. You can skip it.

0 of 10 completed

Progress saved in this browser.

Problem 1 of 10

Problem 1 The linked records

The complete records for $f$ are $f(1)=-4$ and $f(3)=2$ . The complete records for $g$ are $g(-2)=3$ and $g(1)=7$ . Find $(f\circ g)(-2)$ .
1. Hint 1
  The output of the inner rule becomes the input of the outer rule.
2. Hint 2
  Look up the inner result, then use that result as an input in the other record.
Answer

$2$ .

Full solution

The inner rule gives
$g(-2)=3.$
This value belongs to the domain of $f$ , and
$f(3)=2.$
Therefore
$(f\circ g)(-2)=2.$

Answer

$2$ .

Key idea

Composition follows one record’s output into the next record’s input.
Problem 1: Mark complete
Problem 2 A pair of totals

At input $5$ , two functions satisfy $(f+g)(5)=14$ and $(f-g)(5)=-4$ . Find $(fg)(5)$ .
1. Hint 1
  The sum and difference together determine the two individual outputs.
2. Hint 2
  Add the two given equations to find twice the output of $f$ , then recover the output of $g$ .
Answer

$45$ .

Full solution

Let $u=f(5)$ and $v=g(5)$ .
Then $u+v=14$ and $u-v=-4$ .
Adding gives
$2u=10,$
so $u=5$ .
The sum then gives
$v=14-5,$
hence $v=9$ .

The product function multiplies the outputs at this same input.
$(fg)(5)=5\times9.$
Thus
$(fg)(5)=45.$
Checking $5+9=14$ and $5-9=-4$ verifies the recovered outputs.

Answer

$45$ .

Key idea

A function product uses the individual outputs, which can be recovered from their sum and difference.
Problem 2: Mark complete
Problem 3 A quotient’s domain

Let $f(x)=\sqrt{x+2}$ on its real domain and $g(x)=x+2$ on all real numbers. Give a formula and the domain for $f/g$ .
1. Hint 1
  The input must be accepted by both rules, and the divisor output must be nonzero.
2. Hint 2
  Combine the square-root condition with the condition that $x+2$ is not zero.
Answer

$\left(\frac fg\right)(x)=\frac{\sqrt{x+2}}{x+2}$ , or equivalently $\frac1{\sqrt{x+2}}$ ; domain $x>-2$ .

Full solution

Dividing the two outputs gives
$\left(\frac fg\right)(x)=\frac{\sqrt{x+2}}{x+2}.$
The numerator requires $x+2\ge0$ , while the divisor requires $x+2\ne0$ .
Together these give
$x>-2.$
Every such input passes both conditions.
On that domain $x+2$ is positive, so the same rule can equally be written $\frac1{\sqrt{x+2}}$ , and the two forms agree at every allowed input.

Answer

$\left(\frac fg\right)(x)=\frac{\sqrt{x+2}}{x+2}$ , or equivalently $\frac1{\sqrt{x+2}}$ ; domain $x>-2$ .

Key idea

A function quotient combines the original domain requirements with the requirement that its divisor output be nonzero.
Problem 3: Mark complete
Problem 4 A rectangular sample

A rectangular sample has side lengths $u(t)=t+2$ and $v(t)=5-t$ centimeters, where $0\le t\le3$ . Give expanded polynomial formulas for $(u+v)(t)$ , $(u-v)(t)$ , and $(uv)(t)$ , and identify which one gives the sample’s area.
1. Hint 1
  Each combination uses the two side lengths at the same setting.
2. Hint 2
  Keep parentheses when subtracting the second rule, and multiply the full expressions for the product.
Answer

$(u+v)(t)=7$ ; $(u-v)(t)=2t-3$ ; $(uv)(t)=-t^2+3t+10$ , the area in square centimeters; $0\le t\le3$ .

Full solution

Adding and subtracting at the same input gives
$\begin{aligned}(u+v)(t)&=(t+2)+(5-t)\\(u+v)(t)&=7\\(u-v)(t)&=(t+2)-(5-t)\\(u-v)(t)&=2t-3\end{aligned}$

The product gives area in square centimeters.
$\begin{aligned}(uv)(t)&=(t+2)(5-t)\\(uv)(t)&=-t^2+3t+10\end{aligned}$
The model retains domain $0\le t\le3$ , on which both sides are positive.
At $t=0$ , the sides are $2$ and $5$ , giving sum $7$ , difference $-3$ , and area $10$ , which checks all three formulas.

Answer

$(u+v)(t)=7$ ; $(u-v)(t)=2t-3$ ; $(uv)(t)=-t^2+3t+10$ , the area in square centimeters; $0\le t\le3$ .

Key idea

Adding, subtracting, and multiplying two functions combines their outputs at one shared input.
Problem 4: Mark complete
Problem 5 The missing outer rule

Let $g(x)=x-3$ for all real $x$ . A function $f$ satisfies $f(g(x))=2x^2-12x+23$ for every real $x$ . Find $f(t)$ as an expanded polynomial.
1. Hint 1
  Rewrite the input of $f$ as a single new variable.
2. Hint 2
  If $t=x-3$ , express $x$ in terms of $t$ and substitute that expression into the given output.
Answer

$f(t)=2t^2+5$ .

Full solution

Set $t=x-3$ , so $x=t+3$ , and every real $t$ occurs this way.
Then $f(t)$ is the given output with $x$ replaced by $t+3$ :
$2(t+3)^2-12(t+3)+23.$
Expand the squared term first:
$2(t+3)^2=2t^2+12t+18.$
The other product is
$-12(t+3)=-12t-36.$
The linear terms cancel, and $18-36+23=5$ , so
$f(t)=2t^2+5.$

Check by applying this rule to $g(x)$ .
$f(g(x))=2(x-3)^2+5.$
Expanding recovers $2x^2-12x+23$ , as required.

Answer

$f(t)=2t^2+5$ .

Key idea

A known inner rule can be replaced by a new input variable to reconstruct an unknown outer rule.
Problem 5: Mark complete
Problem 6 Passing two gates

Let $g(x)=\sqrt{x+5}$ on its real domain and $f(u)=1/(u-4)$ for $u\ne4$ . Find a formula and the domain for $f\circ g$ .
1. Hint 1
  An input must survive the inner rule and produce an output accepted by the outer rule.
2. Hint 2
  First require a real root; then find which input makes that root equal the forbidden outer input.
Answer

$(f\circ g)(x)=\frac1{\sqrt{x+5}-4}$ , or equivalently $\frac{\sqrt{x+5}+4}{x-11}$ ; $x\ge-5$ and $x\ne11$ .

Full solution

Substituting the inner output gives
$(f\circ g)(x)=\frac1{\sqrt{x+5}-4}.$
The inner rule requires $x\ge-5$ .
The outer rule also rejects the input for which
$\sqrt{x+5}=4.$
Squaring gives $x+5=16$ , hence $x=11$ .

Therefore the composite accepts exactly $x\ge-5$ with $x\ne11$ .
The endpoint $-5$ produces inner output $0$ , which the outer rule accepts, so it remains included.

Answer

$(f\circ g)(x)=\frac1{\sqrt{x+5}-4}$ , or equivalently $\frac{\sqrt{x+5}+4}{x-11}$ ; $x\ge-5$ and $x\ne11$ .

Key idea

A composite may exclude inputs both before and after the inner rule produces its output.
Problem 6: Mark complete
Problem 7 Two possible orders

Let $f(x)=2-x$ and $g(x)=x^2$ , both on all real numbers. Find every input at which $f(g(x))$ and $g(f(x))$ give the same output.
1. Hint 1
  Compare complete formulas for the two orders.
2. Hint 2
  Set those formulas equal and solve the resulting equation.
Answer

$x=1$ .

Full solution

The two orders give $f(g(x))=2-x^2$ and
$g(f(x))=(2-x)^2.$
Equate them and expand:
$\begin{aligned}2-x^2&=x^2-4x+4\\0&=2x^2-4x+2\\0&=2(x-1)^2\end{aligned}$
Thus $x=1$ is the only solution.

At $x=1$ , each inner rule gives $1$ , and both final outputs are $1$ .
The equality at this input does not make the two composite functions identical.

Answer

$x=1$ .

Key idea

Different composite functions can agree at particular inputs without agreeing everywhere.
Problem 7: Mark complete
Problem 8 A check at zero

Let $f(x)=x+2$ and $g(x)=x^2-x$ , both on all real numbers. A student computes $f(g(0))=2$ and $g(f(0))=2$ , both correct, and concludes from those two results that $f\circ g=g\circ f$ . State what the two computed outputs do and do not establish, then work out formulas for the two orders and say whether the conclusion holds.
1. Hint 1
  Equality of functions requires agreement at every allowed input.
2. Hint 2
  Substitute each whole inner rule into the other function to build both composite formulas, then compare them term by term.
Answer

The two outputs show only that the orders agree at $x=0$ , not that the functions are equal. The conclusion fails: $f(g(x))=x^2-x+2$ while $g(f(x))=x^2+3x+2$ , and these differ at every input except $0$ .

Full solution

Agreement at one input is a check of a single case, since two different composite functions can still share an output there.
So the computed pair establishes only that the two orders match at input $0$ , and it cannot establish that the composite functions are equal.
Settling the conclusion needs the two formulas.

Substituting the whole inner rule each way gives
$f(g(x))=(x^2-x)+2$
and hence $f(g(x))=x^2-x+2$ , while
$g(f(x))=(x+2)^2-(x+2)$
expands to
$g(f(x))=x^2+3x+2.$

Subtracting the first formula from the second gives
$g(f(x))-f(g(x))=4x,$
which is zero only at $x=0$ .
So the orders disagree at every other input, for instance $f(g(1))=2$ against $g(f(1))=6$ , and the student’s conclusion does not hold.

Answer

The two outputs show only that the orders agree at $x=0$ , not that the functions are equal. The conclusion fails: $f(g(x))=x^2-x+2$ while $g(f(x))=x^2+3x+2$ , and these differ at every input except $0$ .

Key idea

A single disagreement disproves equality of functions, while a single agreement does not prove it.
Problem 8: Mark complete
Problem 9 Absolute value and squaring

For all real inputs, let $p(x)=|x|$ and $q(x)=x^2$ . Decide whether $p(q(x))$ and $q(p(x))$ give the same output at every real input, and justify your decision.
1. Hint 1
  Both orders involve the size of the input and its square.
2. Hint 2
  Use the facts that a square is nonnegative and squaring opposite numbers gives the same result.
Answer

They agree; both orders give $x^2$ for every real $x$ .

Full solution

In the first order, the square is already nonnegative, so
$p(q(x))=|x^2|$
simplifies to $x^2$ .
In the other order,
$q(p(x))=|x|^2.$
The numbers $x$ and $|x|$ have the same square, so this also equals $x^2$ .

Both rules accept all real inputs, and the two outputs agree for every one of them.
The two orders therefore give the same output at every real input.

Answer

They agree; both orders give $x^2$ for every real $x$ .

Key idea

Some pairs of functions compose equally in both orders, but their rules must justify the agreement.
Problem 9: Mark complete
Problem 10 Two finite records

The complete pairs for $p$ are $(-1,4)$ , $(0,1)$ , and $(2,3)$ . The complete pairs for $q$ are $(1,-1)$ , $(3,0)$ , and $(4,8)$ . Jo says both composites have the same domain. Is Jo correct? Give each composite’s domain.
1. Hint 1
  Follow each allowed inner input through both records.
2. Hint 2
  An inner output must occur among the other record’s input values before the composition can continue.
Answer

No; domain of $p\circ q$ is $\{1,3\}$ , while domain of $q\circ p$ is $\{-1,0,2\}$ .

Full solution

For $p\circ q$ , inputs $1$ and $3$ pass through $q$ to $-1$ and $0$ , both accepted by $p$ .
But
$q(4)=8,$
and $p$ does not accept input $8$ .
Thus the domain of $p\circ q$ is $\{1,3\}$ .

For $q\circ p$ , the inner outputs are
$\begin{aligned}p(-1)&=4\\p(0)&=1\\p(2)&=3.\end{aligned}$
All three outputs are accepted by $q$ , so the domain of $q\circ p$ is $\{-1,0,2\}$ .
The two domains differ, so Jo is incorrect.

Answer

No; domain of $p\circ q$ is $\{1,3\}$ , while domain of $q\circ p$ is $\{-1,0,2\}$ .

Key idea

Reversing a composition may change which inputs can pass through both functions.
Problem 10: Mark complete

← Back to lesson