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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: Core (core-course level)

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Problem 1 of 10

Problem 1 Three reversed records

A function consists of the pairs $\{(-2,4),(0,1),(3,-1)\}$ , with its codomain equal to its range. Write its inverse as a set of ordered pairs.
1. Hint 1
  An inverse exchanges each input and its output.
2. Hint 2
  Check that the original outputs are all different before reversing the pairs.
Answer

$\{(4,-2),(1,0),(-1,3)\}$ .

Full solution

The outputs $4,1,-1$ are distinct, so reversing them creates no repeated input.
Swapping each pair gives
$\{(4,-2),(1,0),(-1,3)\}.$

Following any original pair and then its reversed pair returns the starting value, and the reverse order does the same.

Answer

$\{(4,-2),(1,0),(-1,3)\}$ .

Key idea

Distinct original outputs make the reversed relation an inverse function.
Problem 1: Mark complete
Problem 2 A reciprocal candidate

Let $f(x)=\frac1{x-1}$ for $x\ne1$ , with range all $u\ne0$ . Verify whether $g(u)=1+\frac1u$ for $u\ne0$ , with range all $x\ne1$ , is its inverse.
1. Hint 1
  An inverse must undo the rule in both orders on the stated domains.
2. Hint 2
  In one order subtract the added $1$ ; in the other take the reciprocal of a reciprocal.
Answer

Yes; $g(f(x))=x$ for $x\ne1$ , and $f(g(u))=u$ for $u\ne0$ .

Full solution

For $x\ne1$ , the value $f(x)$ is nonzero, so
$\begin{aligned}g(f(x))&=1+(x-1)\\&=x\end{aligned}$

For $u\ne0$ , $g(u)-1=1/u$ is nonzero, so $g(u)$ belongs to the domain of $f$ and
$\begin{aligned}f(g(u))&=\frac1{1/u}\\&=u\end{aligned}$
Both required identities hold on their full domains.

Answer

Yes; $g(f(x))=x$ for $x\ne1$ , and $f(g(u))=u$ for $u\ne0$ .

Key idea

An inverse check verifies both compositions and that their intermediate values are allowed.
Problem 2: Mark complete
Problem 3 A reading from the graph

The figure shows the entire graph of a one-to-one function $f$ , whose codomain is its range. Read $f^{-1}(4)$ .

The graph of $f$ .

Text description of this figure

A grid with the horizontal axis labeled x running from -4 to 5 and the vertical axis labeled y running from -3 to 7, gridlines and number labels at every whole number, and the origin labeled 0. The graph labeled f is two straight segments: one from the filled point (-3, -2) to the filled point (1, 0), and one from (1, 0) to the filled point (4, 6). No horizontal guide line, inverse graph, or coordinate label at any requested input is drawn.
1. Hint 1
  The inverse asks which original input produced the given output.
2. Hint 2
  Locate height $4$ on the graph and read its horizontal coordinate.
Answer

$f^{-1}(4)=3$ .

Full solution

On the rising segment from $(1,0)$ to $(4,6)$ , the slope is $2$ , so a rise of $4$ corresponds to a run of $2$ .
Thus height $4$ occurs at input $3$ .
$f(3)=4$

Because the function is one-to-one, this is the unique input producing $4$ , so
$f^{-1}(4)=3.$

Answer

$f^{-1}(4)=3$ .

Key idea

An inverse value is read by starting at an original output and locating its unique input.
Problem 3: Mark complete
Problem 4 A decreasing root

Let $f(t)=4-\sqrt{t+2}$ on $t\ge-2$ , with codomain equal to its range. Find the inverse formula, its domain and its range, and verify the formula in both orders.
1. Hint 1
  Track the algebraic restriction that isolating the square root places on the new input, before computing what that restriction is.
2. Hint 2
  Isolate the nonnegative square root before squaring the equation.
Answer

$f^{-1}(u)=(4-u)^2-2$ ; domain $u\le4$ ; range $[-2,\infty)$ .

Full solution

Writing $u=4-\sqrt{t+2}$ gives $\sqrt{t+2}=4-u$ , requiring $u\le4$ .
Squaring yields
$t=(4-u)^2-2.$
Every $u\le4$ produces an allowed $t\ge-2$ .

In the first order, the inverse applied to $f(t)$ gives
$(\sqrt{t+2})^2-2=t.$
In the other order, the root is $\sqrt{(4-u)^2}=\lvert4-u\rvert=4-u$ because $u\le4$ , so
$f(f^{-1}(u))=u.$

Answer

$f^{-1}(u)=(4-u)^2-2$ ; domain $u\le4$ ; range $[-2,\infty)$ .

Key idea

The sign condition from isolating a square root becomes part of the inverse domain.
Problem 4: Mark complete
Problem 5 A counted order

A shop offers bundles of $n$ items for $C(n)=3n+2$ dollars, where $n\in\{0,1,2,3\}$ and the codomain is $\{2,5,8,11\}$ . Find the inverse rule with its domain, then determine whether an order priced at 7 dollars can be decoded by that inverse.
1. Hint 1
  The inverse inputs are the prices actually produced by allowed item counts.
2. Hint 2
  Solve the price formula for $n$ , but retain the finite input list for the inverse.
Answer

$C^{-1}(p)=\frac{p-2}{3}$ for $p\in\{2,5,8,11\}$ ; 7 dollars is outside its domain.

Full solution

Solve the equation for the count.
$\begin{aligned}p&=3n+2\\n&=\frac{p-2}{3}\end{aligned}$

Different allowed counts give different prices, so the inverse exists on $\{2,5,8,11\}$ .
The number $7$ is not one of these prices; the formula would give $\frac53$ , outside the allowed count set, so the inverse is undefined there.

Answer

$C^{-1}(p)=\frac{p-2}{3}$ for $p\in\{2,5,8,11\}$ ; 7 dollars is outside its domain.

Key idea

An inverse formula must retain any discrete restrictions imposed by the original situation.
Problem 5: Mark complete
Problem 6 Two halves of a scaled rule

The rule $f(x)=\lvert2x-6\rvert$ on $\mathbb R$ is not one-to-one. Give two different interval restrictions that retain the full output range $[0,\infty)$ and make the rule invertible. Give the inverse for each restriction.
1. Hint 1
  The two sides of the point $x=3$ produce the same distances scaled by the same factor.
2. Hint 2
  Keep one complete side at a time, then resolve the absolute value using that side condition before dividing by the scale factor.
Answer

On $[3,\infty)$ : $f^{-1}(u)=\frac{u+6}2$ ; on $(-\infty,3]$ : $f^{-1}(u)=\frac{6-u}2$ . Both inverse domains are $[0,\infty)$ .

Full solution

For $x\ge3$ , the rule is $u=2x-6$ , so
$x=\frac{u+6}2.$
For $x\le3$ , the rule is $u=6-2x$ , so
$x=\frac{6-u}2.$
Each half reaches every $u\ge0$ once.

In each case substitution recovers $u$ , and the output of the inverse belongs to its chosen half-line.
These are different restricted functions with different inverses, although their original formulas match.

Answer

On $[3,\infty)$ : $f^{-1}(u)=\frac{u+6}2$ ; on $(-\infty,3]$ : $f^{-1}(u)=\frac{6-u}2$ . Both inverse domains are $[0,\infty)$ .

Key idea

A domain restriction selects which input will be recovered from each repeated output.
Problem 6: Mark complete
Problem 7 A rule used twice

Let $f(x)=\frac{x+3}{2x-1}$ for $x\ne\frac12$ , whose outputs also avoid $\frac12$ . Determine whether $f$ is its own inverse and verify your conclusion.
1. Hint 1
  Solving for the original input can reveal whether the same formula reverses the rule.
2. Hint 2
  In the double application, simplify its numerator and denominator separately.
Answer

Yes; $f^{-1}(x)=\frac{x+3}{2x-1}$ on $x\ne\frac12$ .

Full solution

The output cannot equal $\frac12$ , because $2(x+3)=2x-1$ is impossible.
Therefore the second application is defined.
If $v=f(x)$ , then
$\begin{aligned}v+3&=\frac{7x}{2x-1}\\2v-1&=\frac7{2x-1}\end{aligned}$

The second expression is nonzero on the domain.
Their quotient is $x$ , so $f(f(x))=x$ throughout
$x\ne\frac12.$
Since both compositions are the same double application, both inverse identities hold.

Answer

Yes; $f^{-1}(x)=\frac{x+3}{2x-1}$ on $x\ne\frac12$ .

Key idea

A function whose second application returns each allowed input is its own inverse.
Problem 7: Mark complete
Problem 8 Two claimed reversals

Let $f:A\to B$ be one-to-one with range $B$ . Two functions $g,h:B\to A$ both claim to be inverses of $f$ . Explain why they cannot disagree at any input $b\in B$ .
1. Hint 1
  Each inverse must return an input that $f$ sends to $b$ .
2. Hint 2
  Use one-to-oneness to compare the two proposed original inputs.
Answer

$g(b)=h(b)$ for every $b\in B$ ; the inverse is unique.

Full solution

The inverse identities require
$\begin{aligned}f(g(b))&=b\\f(h(b))&=b\end{aligned}$

Thus the two inputs $g(b)$ and $h(b)$ have the same image under the one-to-one function $f$ .
They must be equal.
This holds for every $b$ , and the two functions share domain $B$ and codomain $A$ , so $g=h$ .

Answer

$g(b)=h(b)$ for every $b\in B$ ; the inverse is unique.

Key idea

One-to-oneness ensures that each output has at most one original input for an inverse to return.
Problem 8: Mark complete
Problem 9 A student's verification

Let $f(x)=\lvert x-1\rvert$ map $\mathbb R$ onto $[0,\infty)$ , and let $g(u)=u+1$ map $[0,\infty)$ into $\mathbb R$ . A student checks $f(g(u))=u$ and declares $g=f^{-1}$ . Is that conclusion valid? Explain by checking the other order.
1. Hint 1
  A one-sided check is not the definition of an inverse.
2. Hint 2
  Compare $g(f(x))$ with $x$ for an input below $1$ .
Answer

No; $g(f(x))=\lvert x-1\rvert+1$ , which differs from $x$ for $x<1$ .

Full solution

For $u\ge0$ , $f(g(u))=\lvert u\rvert=u$ , so the reported check is correct.
But the other order gives
$g(f(x))=\lvert x-1\rvert+1.$

For $x=0$ , this is $2$ , not $0$ .
Thus $g$ does not undo $f$ on its entire domain.
The original rule gives the same output at $0$ and $2$ , so it is not one-to-one and has no inverse with that full domain.

Answer

No; $g(f(x))=\lvert x-1\rvert+1$ , which differs from $x$ for $x<1$ .

Key idea

A function that chooses one original input does not undo a rule that merged several original inputs.
Problem 9: Mark complete
Problem 10 A horizontal segment

The figure shows a function on $[-3,3]$ , with codomain its range. A student claims that swapping every coordinate pair in its graph gives another function. Is the claim correct? Explain what happens to the horizontal segment.

The graph of the function.

Text description of this figure

A grid with the horizontal axis labeled x running from -4 to 4 and the vertical axis labeled y running from -2 to 4, gridlines and number labels at every whole number, and the origin labeled 0. The graph is three straight segments joining the filled points (-3, -1), (-1, 1), (2, 1), and (3, 3) in order. The middle segment, from (-1, 1) to (2, 1), is horizontal. No swapped relation, guide line, or inverse annotation is drawn.
1. Hint 1
  Repeated original outputs become repeated inputs after the coordinates are swapped.
2. Hint 2
  Observe how many original inputs share the height of the horizontal segment.
Answer

No; the horizontal segment becomes the vertical segment from $(1,-1)$ to $(1,2)$ .

Full solution

The original graph includes $(x,1)$ for every $x\in[-1,2]$ .
Swapping the coordinates gives
$\{(1,x):-1\le x\le2\}.$

That is a vertical segment, assigning many outputs to the input $1$ .
The original fails the horizontal line test, and its swapped relation fails the vertical line test, so the swap is not a function.

Answer

No; the horizontal segment becomes the vertical segment from $(1,-1)$ to $(1,2)$ .

Key idea

A horizontal segment reveals repeated outputs that prevent an inverse relation from being a function.
Problem 10: Mark complete

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