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Graphs of 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 Reading one inverse value

The graph shows the complete one-to-one function $f$ . Find $f^{-1}(3)$ .

The complete graph of $f$ , with the diagonal $y = x$ .

Text description of this figure

A square coordinate grid. The horizontal x axis and the vertical y axis each run from negative 4 to 6, with tick marks, number labels and gridlines at every whole number, equal unit lengths on both axes, and the origin labeled 0. A straight segment falls from a filled dot at the point (negative 3, 4) to a filled dot at the point (5, 0), passing through the grid intersections (negative 1, 3), (1, 2) and (3, 1). A dashed diagonal line labeled y = x runs from the lower left corner of the grid to the upper right corner. No coordinates are printed beside the dots, and no other curve is drawn.
1. Hint 1
  An input of $f^{-1}$ is a height that the graph of $f$ actually reaches.
2. Hint 2
  Find the point of the graph at height $3$ , then read its first coordinate.
Answer

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

Full solution

The graph of $f$ is the segment falling from $(-3,4)$ to $(5,0)$ , and it drops one unit for every two units to the right.
Two steps right of $(-3,4)$ the height has fallen by one, so the segment passes through the grid point $(-1,3)$ and $f(-1)=3$ .
Reflecting that point across $y=x$ gives $(3,-1)$ on the graph of $f^{-1}$ , which records
$f^{-1}(3)=-1.$
Each height on this falling segment occurs once, so no other input gives the height $3$ .

Answer

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

Key idea

An inverse value is read off the original graph by finding the required height and taking the input that reached it.
Problem 1: Mark complete
Problem 2 A graph with two pieces

The graph shows the complete one-to-one function $f$ . Is $0$ in the domain of $f^{-1}$ ?

The complete graph of $f$ .

Text description of this figure

A square coordinate grid. The horizontal x axis and the vertical y axis each run from negative 5 to 4, with tick marks, number labels and gridlines at every whole number, equal unit lengths on both axes, and the origin labeled 0. The graph is two separate straight segments, each with a filled dot at both ends: one rises from (negative 4, negative 4) to (negative 2, negative 2), and the other rises from (1, 1) to (3, 3). Between them the grid is empty, and no coordinates, extra points or other lines are drawn.
1. Hint 1
  An inverse input must be an output actually reached by the original graph.
2. Hint 2
  Look for a point of the original graph at height zero.
Answer

No; $0$ is not in the domain of $f^{-1}$ .

Full solution

The first piece reaches heights from $-4$ through $-2$ , and the second reaches heights from $1$ through $3$ .
Neither reaches height zero, and the pieces are not connected.
The graph has no point of the form $(x,0)$ .
Zero is therefore absent from the inverse domain, so $f^{-1}(0)$ is undefined.

Answer

No; $0$ is not in the domain of $f^{-1}$ .

Key idea

A gap in the original range becomes a gap in the inverse’s input set.
Problem 2: Mark complete
Problem 3 The two coordinate spans

The figure shows the complete one-to-one function $g$ . How long is the horizontal span of its inverse graph, measured in coordinate units?

The complete graph of $g$ , with the diagonal $y = x$ .

Text description of this figure

A square coordinate grid. The horizontal x axis and the vertical y axis each run from negative 4 to 6, with tick marks, number labels and gridlines at every whole number, equal unit lengths on both axes, and the origin labeled 0. Two joined straight segments rise from a filled dot at the point (negative 2, negative 2), through a filled corner dot at (0, 1), to a filled dot at (3, 4). A dashed diagonal line labeled y = x runs from the lower left corner of the grid to the upper right corner. No coordinates are printed beside the dots, and no other curve is drawn.
1. Hint 1
  The inverse’s horizontal spread comes from the original graph’s vertical spread.
2. Hint 2
  Read the highest and lowest original outputs and subtract them.
Answer

$6$ coordinate units.

Full solution

The original outputs run from $-2$ to $4$ .
After reflecting across $y=x$ , those become the horizontal endpoints of the inverse domain.
Its span is therefore
$4-(-2)=6.$
The original horizontal span is a different measurement and becomes the inverse’s vertical span.

Answer

$6$ coordinate units.

Key idea

Reflecting a graph across $y=x$ trades its horizontal and vertical extents, so the inverse’s horizontal span is the original’s vertical span.
Problem 3: Mark complete
Problem 4 A descending broken line

The graph shows the whole function $f$ . Draw $f^{-1}$ on the same grid, give its domain and range, and identify the marked corner’s image.

The whole graph of $f$ , with its corner marked $C$ and the diagonal $y = x$ .

Text description of this figure

A square coordinate grid. The horizontal x axis and the vertical y axis each run from negative 4 to 6, with tick marks, number labels and gridlines at every whole number, equal unit lengths on both axes, and the origin labeled 0. Two joined straight segments fall from a filled dot at the point (negative 3, 5), through a filled corner dot at (1, 1) that carries the letter C, to a filled dot at (3, 0). A dashed diagonal line labeled y = x runs from the lower left corner of the grid to the upper right corner. No coordinates are printed, and no reflected curve is drawn.
1. Hint 1
  A decreasing graph can still be one-to-one when each height occurs once.
2. Hint 2
  Swap the coordinates of every endpoint and corner, then reflect the entire connecting segments.
Answer

Inverse joins $(0,3)$ , $(1,1)$ , and $(5,-3)$ . Domain $0\le x\le5$ ; range $-3\le y\le3$ . Corner image: $(1,1)$ .

Full solution

The original points $(-3,5)$ , $(1,1)$ , and $(3,0)$ reverse to $(5,-3)$ , $(1,1)$ , and $(0,3)$ .
Join those reversed points with the corresponding solid segments.
Each original height occurs once, so the reflected graph is a function.

The original range from zero to five becomes the inverse domain
$0\le x\le5.$
The original domain from $-3$ to $3$ becomes the inverse range
$-3\le y\le3.$
The marked corner $(1,1)$ lies on the diagonal and therefore stays at the same point.

Answer

Inverse joins $(0,3)$ , $(1,1)$ , and $(5,-3)$ . Domain $0\le x\le5$ ; range $-3\le y\le3$ . Corner image: $(1,1)$ .

Key idea

A decreasing one-to-one graph reflects into an inverse graph, and a diagonal corner is fixed by that reflection.
Problem 4: Mark complete
Problem 5 A selected branch

The figure shows the parabola $y=(x+1)^2$ . Restrict its domain so that the restricted piece has an inverse function whose range is $-3\le y\le-1$ . State the retained domain, and give the inverse rule with its own domain and range.

The parabola $y = (x+1)^2$ and the diagonal $y = x$ .

Text description of this figure

A square coordinate grid. The horizontal x axis runs from negative 5 to 5 and the vertical y axis runs from negative 1 to 9, with tick marks, number labels and gridlines at every whole number, equal unit lengths on both axes, and the origin labeled 0. A smooth upward parabola labeled y = (x + 1) squared has its lowest point at (negative 1, 0) and rises on both sides, passing through (negative 2, 1), (0, 1), (negative 3, 4) and (1, 4), and reaching the top edge of the grid at (negative 4, 9) and (2, 9), where an arrowhead on each branch shows that the curve continues. A dashed diagonal line labeled y = x runs from (negative 1, negative 1) up to (5, 5). No part of the parabola is highlighted and no other curve is drawn.
1. Hint 1
  The range of an inverse is the domain of the piece it was made from.
2. Hint 2
  Swap the endpoint coordinates, then decide whether $y+1$ is the positive or the negative square root of $x$ .
Answer

Retain $-3\le x\le-1$ . Inverse: $y=-\sqrt{x}-1$ for $0\le x\le4$ , with range $-3\le y\le-1$ .

Full solution

The range of the inverse is the domain of the retained piece, so the inputs to keep are exactly
$-3\le x\le-1.$
On those inputs $x+1$ runs from $-2$ up to $0$ , and squaring numbers that are never positive reverses their order, so the piece falls steadily and each height occurs once.
Its heights run from $4$ at $x=-3$ down to $0$ at $x=-1$ .

Swapping coordinates, the inverse satisfies $x=(y+1)^2$ with
$-3\le y\le-1.$
Since $y+1$ is never positive there, $y+1=-\sqrt{x}$ , giving
$y=-\sqrt{x}-1.$
Its domain is the retained piece’s range, $0\le x\le4$ , and its range is the retained domain,
$-3\le y\le-1.$
At the endpoints, $-\sqrt{0}-1=-1$ and $-\sqrt{4}-1=-3$ , matching the retained inputs.

Answer

Retain $-3\le x\le-1$ . Inverse: $y=-\sqrt{x}-1$ for $0\le x\le4$ , with range $-3\le y\le-1$ .

Key idea

A required inverse range fixes which piece to keep, and the branch left of the vertex, where the squared quantity is non-positive, inverts with the negative square root.
Problem 5: Mark complete
Problem 6 The level segment

The graph shows all of $f$ on $-3\le x\le4$ . Restrict the domain to $c\le x\le4$ . Find the smallest $c$ for which the restricted graph has an inverse function, and describe the inverse graph with its endpoints.

The whole graph of $f$ on $-3 \le x \le 4$ , with the diagonal $y = x$ .

Text description of this figure

A square coordinate grid. The horizontal x axis and the vertical y axis each run from negative 4 to 6, with tick marks, number labels and gridlines at every whole number, equal unit lengths on both axes, and the origin labeled 0. Three joined straight segments carry filled dots at the points (negative 3, negative 2), (0, 2), (2, 2) and (4, 5): the first segment rises, the second is level at height 2, and the third rises again. A dashed diagonal line labeled y = x runs from the lower left corner of the grid to the upper right corner. No boundary, retained piece or reflected curve is marked.
1. Hint 1
  A level segment repeats its height at different inputs, and reflection turns that into a vertical segment.
2. Hint 2
  Keep the right endpoint as required and move the left boundary until every remaining height is used once.
Answer

$c=2$ ; inverse is the segment from $(2,2)$ to $(5,4)$ , with both endpoints included.

Full solution

Between inputs zero and two, the original graph remains at height two.
Any $c<2$ retains at least two distinct inputs with that height, so the reflected graph fails the vertical line test.

At $c=2$ , only the final strictly rising segment survives, from $(2,2)$ to $(4,5)$ .
Each height now occurs once.
Its reflection is the segment from $(2,2)$ to $(5,4)$ , including both endpoints.
Thus $c=2$ is both sufficient and the smallest possible boundary.

Answer

$c=2$ ; inverse is the segment from $(2,2)$ to $(5,4)$ , with both endpoints included.

Key idea

Removing a level stretch can repair repeated outputs before the graph is reflected into an inverse.
Problem 6: Mark complete
Problem 7 The adjusted line

The graph shows the complete function $f$ . Let $g(x)=2f(x)$ on the same domain. State the domain and range of $g^{-1}$ , and find $g^{-1}(6)$ .

The complete graph of $f$ , with the diagonal $y = x$ .

Text description of this figure

A square coordinate grid. The horizontal x axis and the vertical y axis each run from negative 1 to 11, with tick marks, number labels and gridlines at every whole number, equal unit lengths on both axes, and the origin labeled 0. A straight segment labeled f rises from a filled dot at the point (0, 1) to a filled dot at the point (2, 5). A dashed diagonal line labeled y = x runs from the lower left corner of the grid to the upper right corner. No scaled segment, reflected segment or coordinate label is drawn.
1. Hint 1
  The graph to be inverted first has its original heights doubled.
2. Hint 2
  Scale the original endpoints, then exchange their coordinates; a target inverse input is a height on the scaled graph.
Answer

Domain $2\le x\le10$ ; range $0\le y\le2$ ; $g^{-1}(6)=1$ .

Full solution

The original segment has endpoints $(0,1)$ and $(2,5)$ .
Vertical scaling gives endpoints $(0,2)$ and $(2,10)$ for $g$ .
Swapping gives the inverse endpoints $(2,0)$ and $(10,2)$ , which are the two ends of the inverse segment.

The original line has slope two, so its height at input one is three.
After scaling, $g(1)=6$ , which reverses to
$g^{-1}(6)=1.$
The scaled range from two through ten is the inverse domain; the unchanged original domain from zero through two is the inverse range.

Answer

Domain $2\le x\le10$ ; range $0\le y\le2$ ; $g^{-1}(6)=1$ .

Key idea

When a graph is transformed before inversion, reverse the transformed pairs and their resulting spans.
Problem 7: Mark complete
Problem 8 Different endpoint heights

The graph has different heights at its two endpoints. Bo says that is enough to ensure its reflection across $y=x$ is a function. Is Bo correct? Give two reflected points that settle the question.

The graph and the diagonal $y = x$ .

Text description of this figure

A square coordinate grid. The horizontal x axis and the vertical y axis each run from negative 2 to 5, with tick marks, number labels and gridlines at every whole number, equal unit lengths on both axes, and the origin labeled 0. Two joined straight segments carry filled dots at the points (negative 1, 0), (1, 4) and (4, 1): the first rises steeply from the left end to the corner, and the second falls more gently to the right end. A dashed diagonal line labeled y = x runs from the lower left corner of the grid to the upper right corner. No coordinates, test lines or reflected points are drawn.
1. Hint 1
  Different endpoint heights do not rule out a repeated height inside the graph.
2. Hint 2
  Find two original points at the same height and exchange both pairs of coordinates.
Answer

No; the reflection contains $(2,0)$ and $(2,3)$ , giving two outputs at input $2$ .

Full solution

The original graph contains $(0,2)$ and $(3,2)$ , even though its endpoint heights differ.
Reflection exchanges coordinates, producing $(2,0)$ and $(2,3)$ .

These two reflected points have the same input two but distinct outputs zero and three.
A vertical line at input two meets the reflection twice, so the reflection is not a function.
Bo’s endpoint test is insufficient.

Answer

No; the reflection contains $(2,0)$ and $(2,3)$ , giving two outputs at input $2$ .

Key idea

Different endpoint heights do not show that a graph passes the horizontal line test, so they do not guarantee an inverse function.
Problem 8: Mark complete
Problem 9 A stretch of inverse inputs

The complete graph of a function $f$ is the straight segment joining $(-2,9)$ and $(4,-3)$ . Find every input $x$ of $f^{-1}$ for which $f^{-1}(x)>0$ .
1. Hint 1
  An input of $f^{-1}$ is a height on the graph of $f$ , and the matching output is the input that reached that height.
2. Hint 2
  Decide which part of the segment has inputs greater than zero, then read the heights that part covers and test each endpoint.
Answer

$-3\le x<5$ .

Full solution

The segment drops from $9$ to $-3$ across six units, so it falls two units for each unit to the right and its rule is
$f(x)=5-2x$
on
$-2\le x\le4.$
It falls steadily, so each height occurs once and $f^{-1}$ exists.
The heights run from $-3$ up to $9$ , so the domain of $f^{-1}$ is
$-3\le x\le9.$

An inverse output greater than zero means an input of $f$ greater than zero.
Because $f$ falls as its input grows, the inputs $0<x\le4$ cover the heights from $f(4)=-3$ up to, but not including, $f(0)=5$ .
Those heights are the inverse inputs asked for, so the answer is $-3\le x<5$ .
The height $5$ is left out because $f^{-1}(5)=0$ , which is not greater than zero, and the algebraic inverse $f^{-1}(x)=\dfrac{5-x}{2}$ agrees, being positive exactly when $x<5$ .

Answer

$-3\le x<5$ .

Key idea

A condition on an inverse’s outputs becomes a condition on the original graph’s heights, and the boundary height needs its own test.
Problem 9: Mark complete
Problem 10 The three plotted points

The complete graph of $f$ consists of the three plotted points. Kai says reflecting the graph across $y=x$ leaves the whole graph unchanged, even though some individual points move. Is Kai correct? Explain and identify every point that stays fixed.

The three points that make up the graph of $f$ , with the diagonal $y = x$ .

Text description of this figure

A square coordinate grid. The horizontal x axis and the vertical y axis each run from negative 3 to 4, with tick marks, number labels and gridlines at every whole number, equal unit lengths on both axes, and the origin labeled 0. Three filled dots of the same size are plotted at the points (negative 2, 3), (1, 1) and (3, negative 2). Nothing joins them, none of them is labeled or highlighted, and no arrows are drawn. A dashed diagonal line labeled y = x runs from the lower left corner of the grid to the upper right corner.
1. Hint 1
  Compare the whole set of reversed pairs with the original set.
2. Hint 2
  A point itself stays fixed only when its two coordinates are equal.
Answer

Yes; $(-2,3)$ and $(3,-2)$ exchange places, while $(1,1)$ is the only fixed point.

Full solution

The three input values and three output values are all distinct, so the graph represents a one-to-one function.
Reflection exchanges $(-2,3)$ with $(3,-2)$ and leaves $(1,1)$ unchanged.

Thus the set of three plotted points is exactly the same after reflection, confirming Kai’s claim.
The two off-diagonal points move to one another’s positions.
Only $(1,1)$ has equal coordinates and remains individually fixed.

Answer

Yes; $(-2,3)$ and $(3,-2)$ exchange places, while $(1,1)$ is the only fixed point.

Key idea

An inverse graph may equal its original graph as a set while individual off-diagonal points exchange positions.
Problem 10: Mark complete

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