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Inverse 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.

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Problem 1 of 10

Problem 1 Undoing a linear rule

For $f(x)=\frac{3x-1}{4}$ on all real numbers, find $f^{-1}(x)$ and use it to find $f^{-1}(5)$ .
1. Hint 1
  Write the rule as an equation in $x$ and $y$ , then swap the two letters so the old output becomes the new input.
2. Hint 2
  Clear the denominator first, then isolate the new output one operation at a time.
Answer

$f^{-1}(x)=\frac{4x+1}{3}$ , equivalently $\frac43x+\frac13$ ; $f^{-1}(5)=7$ .

Full solution

Write $y=\frac{3x-1}{4}$ and swap the two letters, so the old output becomes the new input.
Then
$x=\frac{3y-1}{4}.$
Multiply by $4$ to clear the denominator:
$4x=3y-1.$
Add $1$ , then divide by $3$ :
$y=\frac{4x+1}{3}.$
Every real input is allowed here, so $f^{-1}(x)=\frac{4x+1}{3}$ on all real numbers.

Substituting the requested value gives
$f^{-1}(5)=\frac{4(5)+1}{3}.$
That is $\frac{21}{3}$ , which equals $7$ .
Checking forward, $f(7)=\frac{3(7)-1}{4}$ , which is $5$ , so the inverse returned the input that produced the output $5$ .

Answer

$f^{-1}(x)=\frac{4x+1}{3}$ , equivalently $\frac43x+\frac13$ ; $f^{-1}(5)=7$ .

Key idea

Swapping the two letters and solving for the new output turns a one-to-one rule into the rule that undoes it.
Problem 1: Mark complete
Problem 2 The zero output

Let $f(x)=\frac{x-1}{x+2}$ for $x\ne-2$ . This rule is one-to-one. Find $f^{-1}(0)$ .
1. Hint 1
  The inverse asks which allowed input produces the output written inside its parentheses.
2. Hint 2
  A fraction with nonzero denominator equals zero when its numerator equals zero.
Answer

$1$ .

Full solution

Solve $f(x)=0$ .
The denominator must remain nonzero, so the numerator condition is
$x-1=0.$
It gives $x=1$ , where the denominator equals $3$ .
Hence $f^{-1}(0)=1$ , and $f(1)=0$ checks the result.

Answer

$1$ .

Key idea

Finding an inverse value means recovering the allowed input that produced a specified output.
Problem 2: Mark complete
Problem 3 The rational rule

For $f(x)=\frac{2x+3}{x-1}$ , with $x\ne1$ , find a single-fraction formula for $f^{-1}(x)$ and state its domain.
1. Hint 1
  Swap the input and output, then collect the terms containing the new output.
2. Hint 2
  After clearing the nonzero denominator, isolate the new output by factoring its coefficient.
Answer

$f^{-1}(x)=\frac{x+3}{x-2}$ ; domain $x\ne2$ .

Full solution

Write $y=(2x+3)/(x-1)$ and swap the variables.
Then
$x=\frac{2y+3}{y-1},$
with $y\ne1$ .
Multiply by this nonzero denominator, then expand and gather every term containing $y$ on one side and everything else on the other:
$\begin{aligned}x(y-1)&=2y+3\\xy-x&=2y+3\\xy-2y&=x+3\\y(x-2)&=x+3\end{aligned}$

If $x=2$ , the last equation would say $0=5$ , so that input is impossible.
For $x\ne2$ , divide by $x-2$ to obtain
$y=\frac{x+3}{x-2}.$
This result never equals $1$ , since its numerator exceeds its denominator by $5$ .
Thus it returns an allowed input of $f$ , completing the reversal.

Answer

$f^{-1}(x)=\frac{x+3}{x-2}$ ; domain $x\ne2$ .

Key idea

Solving the swapped equation reveals both the inverse formula and any output the original rule never attains.
Problem 3: Mark complete
Problem 4 The arrow record

The diagram gives the entire one-to-one function $f$ : its four inputs, its four outputs, and two of its four arrows. The other two arrows have been erased. It is known that $f^{-1}(6)=2$ . Restore both erased arrows, then draw every arrow of the inverse in the lower diagram, whose value columns are drawn but whose arrows are not. State the inverse’s domain and range.

The mapping diagram for $f$ with two arrows erased, above a diagram for $f^{-1}$ whose value columns are drawn but whose arrows are not.

Text description of this figure

Two mapping diagrams, one above the other, with no axes. The upper diagram is titled f. Its left column, labeled Input, holds negative three, two, five and seven from top to bottom, and its right column, labeled Output, holds four, zero, six and one from top to bottom. Two slanted arrows are drawn: one from negative three down to one, and one from five up to four. The inputs two and seven have no arrow leaving them. The lower diagram is titled f inverse. Its left column, labeled Input, holds zero, four, one and six from top to bottom, and its right column, labeled Output, holds two, negative three, five and seven from top to bottom. No arrows are drawn in the lower diagram.
1. Hint 1
  Use the inverse record to recover one of the erased forward pairings.
2. Hint 2
  One input and one output are then still unused, and every listed output is reached exactly once.
3. Hint 3
  Reverse each completed forward arrow, keeping its two endpoint values together: old outputs become inverse inputs.
Answer

Restore $2\mapsto6$ and $7\mapsto0$ . Inverse: $1\mapsto-3$ , $6\mapsto2$ , $4\mapsto5$ , $0\mapsto7$ . Domain $\{0,1,4,6\}$ ; range $\{-3,2,5,7\}$ .

Full solution

The record $f^{-1}(6)=2$ says
$f(2)=6,$
so one erased arrow runs from $2$ to $6$ .
That leaves only the input $7$ unused and only the output $0$ unreached, and every listed output is reached exactly once, so the other erased arrow runs from $7$ to $0$ .
The completed forward pairs are $(-3,1)$ , $(2,6)$ , $(5,4)$ , and $(7,0)$ .

Reverse each arrow to get the inverse pairs $(1,-3)$ , $(6,2)$ , $(4,5)$ , and $(0,7)$ .
Their inputs form the inverse domain $\{0,1,4,6\}$ , and their outputs form its range $\{-3,2,5,7\}$ .
These are exactly the range and the domain of $f$ , traded.

Answer

Restore $2\mapsto6$ and $7\mapsto0$ . Inverse: $1\mapsto-3$ , $6\mapsto2$ , $4\mapsto5$ , $0\mapsto7$ . Domain $\{0,1,4,6\}$ ; range $\{-3,2,5,7\}$ .

Key idea

Restoring the missing pairings and reversing every arrow exchanges a function’s complete domain and range.
Problem 4: Mark complete
Problem 5 A square panel

A square panel has perimeter $p$ centimeters, where $p\ge0$ . Its area in square centimeters is $A(p)=p^2/16$ . Find the inverse rule that recovers perimeter from area, state its domain, and use it for area $144$ square centimeters.
1. Hint 1
  A square’s nonnegative perimeter selects one root when the area rule is reversed.
2. Hint 2
  Swap the area and perimeter roles, clear the denominator, and keep the root permitted by the original domain.
Answer

$A^{-1}(a)=4\sqrt a$ for $a\ge0$ ; perimeter $48$ centimeters.

Full solution

From $a=p^2/16$ , reverse the input-output roles by solving for the original input.
$p^2=16a.$
Since $p\ge0$ ,
$p=4\sqrt a.$
Areas are exactly the nonnegative numbers, so the inverse domain is $a\ge0$ .

At the given area,
$4\sqrt{144}=48.$
The recovered perimeter is $48$ centimeters.
Checking $48^2/16=144$ returns the stated area.

Answer

$A^{-1}(a)=4\sqrt a$ for $a\ge0$ ; perimeter $48$ centimeters.

Key idea

A physical nonnegative quantity determines which root belongs to the inverse of a squared relationship.
Problem 5: Mark complete
Problem 6 The selected inputs

Let $f(x)=(x+2)^2$ with domain $x\le-2$ . Find $f^{-1}(x)$ , its domain and range, and verify both compositions on their respective domains.
1. Hint 1
  The inverse output must belong to the original restricted input set.
2. Hint 2
  After swapping the variables, decide which sign agrees with the restricted inverse outputs.
3. Hint 3
  Use the selected sign when simplifying the square root of a square in the checks.
Answer

$f^{-1}(x)=-2-\sqrt x$ ; domain $x\ge0$ ; range $y\le-2$ . Both inverse compositions return $x$ on their respective domains.

Full solution

Write $y=(x+2)^2$ with $x\le-2$ .
Swapping gives $x=(y+2)^2$ with $y\le-2$ .
Thus $y+2\le0$ , and the correct choice is
$y=-2-\sqrt x.$
The inverse domain is $x\ge0$ , the original range, and its range is $y\le-2$ .

Call the inverse $g$ .
For $x\ge0$ ,
$f(g(x))=(-\sqrt x)^2,$
so $f(g(x))=x$ .
For $x\le-2$ ,
$g(f(x))=-2-\sqrt{(x+2)^2}.$
Here the root is $-(x+2)$ , giving
$g(f(x))=-2+(x+2).$
This equals $x$ , so both directions check.

Answer

$f^{-1}(x)=-2-\sqrt x$ ; domain $x\ge0$ ; range $y\le-2$ . Both inverse compositions return $x$ on their respective domains.

Key idea

A restriction to the negative side of a squared expression determines the negative root in its inverse.
Problem 6: Mark complete
Problem 7 Two meanings in one record

A one-to-one function $f$ has domain and range both equal to $\{-2,2,3\}$ , with $f(-2)=2$ , $f(2)=3$ , and $f(3)=-2$ . Find $f^{-1}(f(3))+1/f(3)$ .
1. Hint 1
  The inverse undoes the forward pairing, while the reciprocal divides one by an output.
2. Hint 2
  Evaluate the two terms separately using the same given output of $f(3)$ .
Answer

$\frac52$ , or $2.5$ .

Full solution

The forward record gives $f(3)=-2$ , and reversing that record gives
$f^{-1}(-2)=3.$
The reciprocal term is instead
$\frac1{f(3)}=-\frac12.$
Therefore the requested sum is
$3-\frac12=\frac52.$
The divisor is nonzero, so the reciprocal is defined.

Answer

$\frac52$ , or $2.5$ .

Key idea

Inverse notation reverses a pairing, while a reciprocal acts arithmetically on the resulting output.
Problem 7: Mark complete
Problem 8 Two complete records

The complete pairs for $f$ are $(0,2)$ and $(1,3)$ . The complete pairs for $g$ are $(2,0)$ , $(3,1)$ , and $(4,0)$ . Exactly one pair must be removed from $g$ to leave two functions that are inverses of each other. State which pair, and explain what that pair did to prevent $f$ and $g$ from being inverses.
1. Hint 1
  Reverse each pair of $f$ , and compare what you get with the pairs of $g$ .
2. Hint 2
  Every allowed input of each function must return to itself; test the other composition at each input of $g$ in turn.
Answer

Remove $(4,0)$ ; the input $4$ is not an output of $f$ , so $f(g(4))=2\ne4$ .

Full solution

Reversing the two pairs of $f$ gives $2\mapsto0$ and $3\mapsto1$ , which are two of the three pairs of $g$ .
Those two behave correctly in both orders: $g(f(0))=0$ and $g(f(1))=1$ , while $f(g(2))=2$ and $f(g(3))=3$ .

The remaining pair $(4,0)$ is the obstruction.
Its input $4$ is not an output of $f$ , so the other order fails there:
$g(4)=0,$
followed by
$f(0)=2.$
Hence $f(g(4))=2$ , which is not $4$ .
That pair also gives $g$ two inputs sharing the output $0$ , so $g$ is not one-to-one.
Removing it leaves the pairs $(2,0)$ and $(3,1)$ , and then every input of each function returns to itself.

Answer

Remove $(4,0)$ ; the input $4$ is not an output of $f$ , so $f(g(4))=2\ne4$ .

Key idea

Two finite records are inverses only when each reverses the other exactly, so one extra pair can break the reverse direction.
Problem 8: Mark complete
Problem 9 A graph with a gap

The graph shows the complete function $f$ , drawn in two separate pieces. Decide whether $f$ has an inverse function, justifying your decision from the graph. If it does, give the inverse’s domain and range.

The complete graph of $f$ .

Text description of this figure

A coordinate grid with equal unit lengths on both axes. The horizontal x-axis is numbered at every whole number from negative five to five, the vertical y-axis at every whole number from negative four to six, the origin is labeled 0, and gridlines run at every whole number. Two solid straight segments are drawn. The first rises from the point (negative 4, negative 3) to the point (negative 2, negative 1). The second rises from the point (1, 2) to the point (4, 5). A filled dot marks each of the four segment ends. Nothing is drawn between the two segments, and no formula, coordinate label or horizontal test line appears.
1. Hint 1
  An inverse needs each output to come from exactly one input; it does not need every possible real output.
2. Hint 2
  Check each piece with horizontal lines, and compare the heights reached by the two pieces.
3. Hint 3
  If no output is repeated, exchange the graph’s horizontal and vertical spans to obtain the inverse’s input and output sets.
Answer

Yes, $f$ has an inverse function. Inverse domain: $-3\le x\le-1$ or $2\le x\le5$ . Inverse range: $-4\le y\le-2$ or $1\le y\le4$ .

Full solution

Each rising segment reaches each of its heights once.
Their output intervals, $-3\le y\le-1$ and $2\le y\le5$ , do not overlap.
Consequently no horizontal line meets the complete graph more than once, so $f$ is one-to-one and has an inverse.

The outputs of $f$ become inverse inputs: $-3\le x\le-1$ or
$2\le x\le5.$
Its inputs become inverse outputs: $-4\le y\le-2$ or
$1\le y\le4.$
The gap simply leaves some values outside these sets, and it blocks nothing, since no output of $f$ is ever repeated.

Answer

Yes, $f$ has an inverse function. Inverse domain: $-3\le x\le-1$ or $2\le x\le5$ . Inverse range: $-4\le y\le-2$ or $1\le y\le4$ .

Key idea

A function with disconnected pieces can have an inverse when no output is repeated across or within its pieces.
Problem 9: Mark complete
Problem 10 A rule that returns

For all real inputs, $f(x)=ax+b$ with $a\ne0$ . It is required that $f$ be its own inverse and that $f(0)=6$ . Find $a$ and $b$ , verify that applying your rule twice returns every real input, and explain why this single check settles both composition requirements.
1. Hint 1
  A function that is its own inverse must return every input after being applied twice.
2. Hint 2
  Use the known value at zero to determine the constant, then substitute the full rule into itself.
3. Hint 3
  The coefficient of the input must be one and the remaining constant must be zero.
Answer

$a=-1$ , $b=6$ ; $f(x)=6-x$ , and $f(f(x))=x$ for every real input, which is both composition checks at once.

Full solution

From $f(0)=6$ , obtain $b=6$ .
Applying the rule twice gives
$f(f(x))=a(ax+6)+6.$
Thus the coefficient of $x$ is $a^2$ , and the constant is $6a+6$ .
Returning every input requires $a^2=1$ and $6a+6=0$ .
The latter gives
$a=-1,$
which also satisfies the first condition.

Therefore $f(x)=6-x$ .
Since the proposed inverse is this same rule, both orders have the same check:
$6-(6-x)=x.$
This holds for every real input, and $f(0)=6$ also holds.

Answer

$a=-1$ , $b=6$ ; $f(x)=6-x$ , and $f(f(x))=x$ for every real input, which is both composition checks at once.

Key idea

Requiring a rule to be its own inverse imposes conditions on every term in its double application.
Problem 10: Mark complete

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