The graph of an inverse
You have the graph of a function and want the graph of its inverse. Do you have to find a formula first, or does the picture already contain the answer?
Where you are 5 in Inverse functions
Why this exists
Finding an inverse from a rule turns a formula into the formula that runs it backwards. Many functions never arrive as a formula. A graph read off an experiment, a curve sketched from a description, a broken line drawn from a few measured points: each is a function you can see but cannot solve. And even when there is a formula, solving it gives no picture of what the inverse looks like.
The graph turns out to need no algebra at all. The inverse undoes f, so every pair it holds is a pair of f turned round. Turning every pair round has one geometric meaning: reflect the whole graph in the line y = x. With that one fact you can draw an inverse you have no formula for, see at a glance whether an inverse exists, read off its domain and range, and predict where the two graphs meet.
Definition
The graph of the inverse f^{-1} is the set of points (b, a) for which (a, b) is a point on the graph of f. In words: take every point of the graph of f and swap its two coordinates.
This comes straight from the definition of the inverse, which reads one fact in two directions:
The left side says (a, b) is on the graph of f. The right side says (b, a) is on the graph of f^{-1}. The two graphs are drawn on the same axes, with x as the input of both.
The line y = x is the graph of the identity function: every point on it has output equal to input. It is the mirror for inverses.
Intuition
Swapping coordinates is a reflection. Plot (1, 2) and (2, 1), or (-1, 0) and (0, -1), on squared paper. Each pair sits like an object and its image in a mirror laid along the diagonal line y = x: the same distance from the line, on opposite sides, with the segment joining them crossing the line square on.
The figure shows \displaystyle f(x) = \frac{1}{4}x^3 - 1, whose graph passes through (2, 1), (0, -1) and (-2, -3), and its inverse, which passes through the turned round points (1, 2), (-1, 0) and (-3, -2). Do this for every point and the whole curve folds over the diagonal. A steep stretch of f becomes a shallow stretch of f^{-1}, because a large rise for a small run turns into a small rise for a large run.
Swapping the coordinates is swapping the axes. Draw a graph on tracing paper, turn the sheet over so the x axis lies where the y axis was and the y axis where the x axis was. The flip is about the line y = x, and what you see through the paper is the graph of the inverse. Inputs have become outputs and outputs inputs, which is the whole idea of an inverse.
Compare the reflections you already know. Reflecting a graph across the x axis keeps each input and changes the sign of the output: (a, b) goes to (a, -b). Reflecting across the y axis sends (a, b) to (-a, b). Reflecting across y = x changes no sign at all; it trades the two coordinates, (a, b) to (b, a). Points on the mirror line, where a = b, do not move.
Results
The mirror. The graph of f^{-1} is the reflection of the graph of f in the line y = x. A point (a, b) is on the graph of f exactly when (b, a) is on the graph of f^{-1}.
Domain and range trade places. The left to right extent of one graph is the bottom to top extent of the other: the domain of f^{-1} is the range of f, and the range of f^{-1} is the domain of f.
When the mirror image is a function. The reflection of the graph of f is the graph of a function exactly when f is one to one, that is, when no horizontal line meets the graph of f more than once.
Lines. If f is a line with slope m \ne 0, then f^{-1} is a line with slope \displaystyle \frac{1}{m}. The y intercept of one is the x intercept of the other.
Where the graphs meet. If f is increasing, the graphs of f and f^{-1} can meet only on the line y = x, so their meeting points are the solutions of f(x) = x.
Self-inverse functions. f is its own inverse exactly when its graph is symmetric about the line y = x.
Why it is true
Why (a, b) and (b, a) are mirror images in y = x. If a = b the point lies on the line and is its own image. Otherwise let P = (a, b) and Q = (b, a). The midpoint of PQ is
whose two coordinates are equal, so it lies on y = x. The slope of PQ is
and the line y = x has slope 1. The product of the slopes is -1, so the two are perpendicular. The line y = x therefore passes through the midpoint of PQ at a right angle: it is the perpendicular bisector of PQ, and that is what it means for Q to be the reflection of P. Since the graph of f^{-1} consists of exactly the swapped points, reflecting every point of f gives every point of f^{-1} and nothing else.
Why the horizontal line test decides. Reflection in y = x sends the horizontal line y = c to the vertical line x = c. So a horizontal line meets the graph of f twice exactly when a vertical line meets the reflected graph twice, and by the vertical line test that is exactly when the reflection fails to be a function.
Why an increasing function meets its inverse only on y = x. Suppose (a, b) lies on both graphs. On f it says f(a) = b; on f^{-1} it says f^{-1}(a) = b, that is, f(b) = a. If a < b, then because f is increasing, f(a) < f(b), which says b < a, a contradiction. If a > b, the same reasoning gives b > a. So a = b and the point is on y = x.
Why lines have reciprocal slopes. Take two points (p, q) and (r, s) on a line with p \ne r. Its slope is \displaystyle \frac{s - q}{r - p}. The mirrored points (q, p) and (s, r) give the inverse the slope \displaystyle \frac{r - p}{s - q}, the reciprocal. Rise and run trade places. The point (0, k) where f crosses the y axis mirrors to (k, 0), where f^{-1} crosses the x axis.
Worked examples
A line and its inverse
Let h(x) = 2x + 5. Find three points on the graph of h^{-1} by mirroring, and check them against the formula for h^{-1}. Compare the slopes and intercepts of the two lines.
SolutionThree points on h: h(0) = 5, h(1) = 7, and h(x) = 0 at \displaystyle x = -\frac{5}{2}. So (0, 5), (1, 7) and \displaystyle \left(-\frac{5}{2}, 0\right) lie on h. Swapping coordinates, (5, 0), (7, 1) and \displaystyle \left(0, -\frac{5}{2}\right) lie on h^{-1}.
For the formula, write y = 2x + 5 and solve for x: \displaystyle x = \frac{y - 5}{2}. Renaming the input,
Check the mirrored points: h^{-1}(5) = 0, \displaystyle h^{-1}(7) = \frac{2}{2} = 1 and \displaystyle h^{-1}(0) = -\frac{5}{2}. All three agree.
The line h has slope 2, y intercept 5 and x intercept \displaystyle -\frac{5}{2}. The line h^{-1} has slope \displaystyle \frac{1}{2}, y intercept \displaystyle -\frac{5}{2} and x intercept 5. The slopes are reciprocals because “up 2 for every 1 across” becomes “up 1 for every 2 across” once rise and run trade places, and the intercepts trade because (0, 5) mirrors to (5, 0).
A cubic, plotted and mirrored
Let \displaystyle f(x) = \frac{1}{4}x^3 - 1. Plot the points of f at x = -2, 0 and 2, mirror them, and confirm that the mirrored points satisfy \displaystyle f^{-1}(x) = \sqrt[3]{4(x + 1)}. State the domain and range of both functions.
Solution\displaystyle f(-2) = \frac{1}{4}(-8) - 1 = -3, f(0) = -1 and \displaystyle f(2) = \frac{1}{4}(8) - 1 = 1. So f passes through (-2, -3), (0, -1) and (2, 1), and its inverse passes through (-3, -2), (-1, 0) and (1, 2). These are the pairs joined in the figure under Intuition.
The formula comes from solving \displaystyle y = \frac{1}{4}x^3 - 1 for x: adding 1 gives \displaystyle y + 1 = \frac{1}{4}x^3, multiplying by 4 gives x^3 = 4(y + 1), and the cube root gives \displaystyle x = \sqrt[3]{4(y + 1)}. Every real number has exactly one cube root, so no case is lost. Check the mirrored points:
A cubic takes every real input and reaches every real output, so f has domain and range all real numbers, and by the trade of domain and range, so does f^{-1}. In the figure the steep climb of f through (2, 1) shows up as the gentle climb of f^{-1} through (1, 2).
Drawing an inverse with no formula
The graph of a function g is the broken line joining (-4, -3), (-1, -2), (1, 1) and (3, 5), in that order. Draw the graph of g^{-1}, give the domain and range of both, and find g^{-1}(3).
SolutionThe graph rises from left to right along every piece, so no horizontal line meets it twice and g has an inverse. A segment reflects to a segment, so reflecting the broken line only needs its corners: (-3, -4), (-2, -1), (1, 1) and (5, 3), joined in the same order.
The graph of g runs from x = -4 to x = 3 and from y = -3 to y = 5, so g has domain [-4, 3] and range [-3, 5]. The mirror image runs from x = -3 to x = 5 and from y = -4 to y = 3: g^{-1} has domain [-3, 5] and range [-4, 3], the two intervals traded.
For g^{-1}(3), you need the input of g whose output is 3. Height 3 is reached on the piece from (1, 1) to (3, 5), which climbs 4 in 2 across, slope 2, so height 3 is reached 1 across from (1, 1), at x = 2. So g(2) = 3 and g^{-1}(3) = 2. On the picture, (3, 2) is on the reflected piece from (1, 1) to (5, 3), halfway along. The corner (1, 1) sits on the mirror line and belongs to both graphs. The two first pieces also cross, at (-2.5, -2.5): the piece of g from (-4, -3) to (-1, -2) meets the line y = x there, and a point on the mirror line is its own image, so it lies on both graphs.
A mirror image that is not a function
Reflect the graph of y = x^2 - 25 in the line y = x. Is the result the graph of a function?
SolutionThe parabola passes through (0, -25), (5, 0), (-5, 0), (6, 11) and (-6, 11). The reflection passes through (-25, 0), (0, 5), (0, -5), (11, 6) and (11, -6): a parabola lying on its side, opening to the right.
It is not the graph of a function. The input 0 is paired with both 5 and -5, so the vertical line x = 0 meets it twice. That failure was already visible in the original: the horizontal line y = 0 meets the parabola at (5, 0) and (-5, 0), because 5 and -5 have the same output. The algebra shows the same thing. Swapping gives x = y^2 - 25, so y^2 = x + 25 and \displaystyle y = \pm\sqrt{x + 25}, and the \pm is two outputs for one input.
Keeping only the right half of the parabola, x \ge 0, removes the repeat: that half reflects to the upper half of the sideways parabola, the graph of \displaystyle \sqrt{x + 25}. Cutting a function down like this is the subject of restricting a domain to get an inverse.
Where a graph meets its inverse
Let \displaystyle p(x) = \frac{x^3 + 6}{7}, whose inverse is \displaystyle p^{-1}(x) = \sqrt[3]{7x - 6}. Find every point where the graphs of p and p^{-1} meet.
SolutionThe function p is increasing: cubing keeps the order of numbers, and adding 6 and dividing by the positive number 7 keep it too. So, by the result above, the graphs meet only on y = x, and the meeting points are the solutions of p(x) = x:
The factors come from trying small divisors of 6: 1 - 7 + 6 = 0 and 8 - 14 + 6 = 0, so x = 1 and x = 2 are roots, and the third factor makes the constant term (-1)(-2)(3) = 6 come out right. The graphs meet at (1, 1), (2, 2) and (-3, -3).
Check on both graphs: \displaystyle p(2) = \frac{14}{7} = 2 and \displaystyle p^{-1}(2) = \sqrt[3]{8} = 2; \displaystyle p(-3) = \frac{-27 + 6}{7} = -3 and \displaystyle p^{-1}(-3) = \sqrt[3]{-27} = -3; and \displaystyle p(1) = \frac{7}{7} = 1. Solving p(x) = p^{-1}(x) directly would mean an equation with a cube root on one side; the mirror reduces it to a cubic with no roots in it.
A function that is its own mirror image
Let \displaystyle r(x) = \sqrt{25 - x^2} for 0 \le x \le 5. Show from its graph that r is its own inverse, and confirm it with algebra.
SolutionThe points of the graph satisfy x^2 + y^2 = 25 with x and y both between 0 and 5: it is the quarter of the circle of radius 5 about the origin that lies in the first quadrant. Swapping the coordinates of such a point gives a point that still satisfies y^2 + x^2 = 25 and still has both coordinates between 0 and 5. So reflecting the quarter circle in y = x gives back the same quarter circle, and r^{-1} = r. Two pairs show it: (3, 4) and (4, 3) are both on the graph, as are (0, 5) and (5, 0).
Algebraically, \displaystyle y = \sqrt{25 - x^2} with y \ge 0 gives y^2 = 25 - x^2, so x^2 = 25 - y^2. Since x \ge 0 on the domain, taking the non-negative root is the only choice: \displaystyle x = \sqrt{25 - y^2}. Renaming, \displaystyle r^{-1}(x) = \sqrt{25 - x^2} for 0 \le x \le 5, the same function. Composing confirms it, using x \ge 0 in the last step:
The function r is decreasing, and its graph shares every point with its inverse, not only the point \displaystyle \left(\frac{5\sqrt{2}}{2}, \frac{5\sqrt{2}}{2}\right) where it crosses y = x. That is no contradiction: the result about meeting only on the mirror line was for increasing functions.
Boundaries
The mirror image is a function only for a one to one function. Every graph can be reflected in y = x, but if some horizontal line meets the original twice, the reflection fails the vertical line test and is not the graph of any function. A horizontal line itself, y = c, reflects to the vertical line x = c, which is why a constant function has no inverse and a slope of 0 has no reciprocal.
The mirror needs equal scales. The line y = x makes a 45^\circ angle with the axes, and the reflection looks like a mirror image, only when one unit is the same length on both axes. On stretched axes the swapped points are still the right points, but the picture no longer looks folded over the diagonal.
f^{-1} is not \displaystyle \frac{1}{f}. The point (-2, -3) on the graph of \displaystyle f(x) = \frac{1}{4}x^3 - 1 mirrors to (-3, -2) on the inverse. The graph of \displaystyle \frac{1}{f} would instead have the point \displaystyle \left(-2, -\frac{1}{3}\right), keeping the input in place. Mirroring trades coordinates; it never takes reciprocals of outputs.
Meeting only on the mirror line needs an increasing function. A decreasing function can meet its inverse off the line y = x. For q(x) = -x^3 the points (1, -1) and (-1, 1) lie on both graphs (Exercise 16), and the quarter circle of Example 6 shares its whole graph with its inverse.
In a real situation the axes carry different quantities. If C(h) is a cost in dollars after h hours, its inverse takes dollars and gives hours, and its graph has dollars across and hours up. Swapping the axes swaps the labels too. Drawing both graphs on one pair of axes and folding over y = x only makes sense when the input and output are measured in the same units, or when they are treated as plain numbers.
Where this goes next
The most important mirror pair in the subject is the exponential function and the logarithm. The graph of a logarithm is drawn by reflecting the graph of an exponential in y = x, and every feature of it, the point (1, 0), the vertical asymptote, the slow climb, is a feature of the exponential turned round. The same move draws the inverse trigonometric functions, each the mirror of a one to one slice of a wave. In calculus the reciprocal slopes of mirrored lines become the derivative of an inverse: the slope of f^{-1} at (b, a) is the reciprocal of the slope of f at (a, b).
Exercises
Points and their mirror images
- The graph of f passes through (3, 7), (-2, 0) and (5, -4). Name three points on the graph of f^{-1}.
- Reflect each point in the line y = x: (6, -1), (0, 4), (-3, -3).
- Show that (1, 6) and (6, 1) are mirror images in y = x by finding the midpoint of the segment joining them and its slope.
- A function g has g(4) = -9. Which point must lie on the graph of g^{-1}, and what is g^{-1}(-9)?
- Let f(x) = x + 2, so (2, 4) is on its graph. Give the point on the graph of f^{-1} that comes from (2, 4), and the point on the graph of \displaystyle \frac{1}{f} with first coordinate 2.
Lines
- Let k(x) = 4x - 8. Find k^{-1}(x), and give the slope, y intercept and x intercept of both lines.
- A line has slope \displaystyle -\frac{3}{5} and passes through (0, 2). Without solving any equation, give the slope of its inverse and one point on it. Then write the inverse as a formula.
- The inverse of a line has slope 6 and passes through (1, -2). Find the original line.
Drawing inverses from points
- The graph of f is the broken line joining (-5, -2), (-1, 0), (3, 1) and (7, 4). Give the corners of the graph of f^{-1}, its domain and range, and \displaystyle f^{-1}\left(\frac{1}{2}\right).
- Let t(x) = x^3 + 4. Plot the points at x = -1, 0 and 1, mirror them, and check that the mirrored points lie on \displaystyle t^{-1}(x) = \sqrt[3]{x - 4}.
- For \displaystyle f(x) = \frac{1}{4}x^3 - 1, find f^{-1}(15) and f^{-1}(-17) by finding a point on f whose output is the given number, then check with \displaystyle f^{-1}(x) = \sqrt[3]{4(x + 1)}.
Mirror images that are not functions
- Reflect the points of y = x^2 + 5 at x = -2, -1, 0, 1 and 2 in the line y = x. Explain from the mirrored points why the reflected graph is not a function.
- For which of these does reflecting the graph in y = x give the graph of a function? (a) y = x^3 (b) y = |x| (c) y = 2x - 1 (d) y = x^4 (e) y = 5
Where a graph meets its inverse
- Let \displaystyle u(x) = \frac{x^3 + 12}{13}, which is increasing. Find every point where the graphs of u and u^{-1} meet.
- Where does the graph of v(x) = 3x - 4 meet the graph of its inverse?
- Let q(x) = -x^3. Find q^{-1}(x), and show that (1, -1), (-1, 1) and (0, 0) lie on both graphs. Why does this not contradict the result about meeting only on y = x?
- Show that \displaystyle d(x) = \frac{9}{x} is its own inverse, and name two pairs of mirror image points on its graph.
Reasoning
- Explain, using the graph, why a line with slope 0 has no inverse function. What does its reflection in y = x look like?
- Explain why every line y = -x + c is its own inverse, first from the graph and then with algebra.
- Selin draws f with one unit across as 1 centimeter and one unit up as 1 millimeter, then swaps the coordinates of several points to draw f^{-1}. Her two curves do not look like mirror images in y = x. Is her inverse wrong? Explain.
- Prove that if the graph of f is symmetric about the line y = x, then f is its own inverse.
- Prove, using two points and their mirror images, that the inverse of a line with slope m \ne 0 has slope \displaystyle \frac{1}{m}.
Answers
- (7, 3), (0, -2) and (-4, 5). Each point of f is turned round.
- (-1, 6), (4, 0) and (-3, -3). The last point is on the mirror line, so it does not move.
- Midpoint \displaystyle \left(\frac{7}{2}, \frac{7}{2}\right), which is on y = x; slope \displaystyle \frac{1 - 6}{6 - 1} = -1, perpendicular to y = x (slope product -1). So y = x is the perpendicular bisector of the segment.
- (-9, 4), and g^{-1}(-9) = 4. g(4) = -9 puts (4, -9) on g, so (-9, 4) is on g^{-1}.
- (4, 2) on f^{-1}, but \displaystyle \left(2, \frac{1}{4}\right) on \displaystyle \frac{1}{f}. The inverse swaps coordinates; \displaystyle \frac{1}{f} keeps the input 2 and takes the reciprocal of the output 4.
- \displaystyle k^{-1}(x) = \frac{x}{4} + 2, from \displaystyle x = \frac{y + 8}{4}. k: slope 4, y intercept -8, x intercept 2. k^{-1}: slope \displaystyle \frac{1}{4}, y intercept 2, x intercept -8.
- Slope \displaystyle -\frac{5}{3}, through (2, 0). The inverse is \displaystyle y = -\frac{5}{3}(x - 2), that is \displaystyle f^{-1}(x) = \frac{10}{3} - \frac{5}{3}x. Check from \displaystyle y = 2 - \frac{3}{5}x: \displaystyle x = \frac{5}{3}(2 - y).
- \displaystyle y = \frac{1}{6}x + \frac{4}{3}. The original has the reciprocal slope \displaystyle \frac{1}{6} and passes through the mirrored point (-2, 1): \displaystyle 1 = \frac{1}{6}(-2) + b gives \displaystyle b = \frac{4}{3}.
- Corners (-2, -5), (0, -1), (1, 3), (4, 7); domain [-2, 4], range [-5, 7]; \displaystyle f^{-1}\left(\frac{1}{2}\right) = 1, because the piece of f from (-1, 0) to (3, 1) reaches height \displaystyle \frac{1}{2} halfway along, at x = 1.
- t passes through (-1, 3), (0, 4), (1, 5); mirrored, (3, -1), (4, 0), (5, 1). Check: \displaystyle \sqrt[3]{3 - 4} = -1, \displaystyle \sqrt[3]{0} = 0, \displaystyle \sqrt[3]{1} = 1.
- f^{-1}(15) = 4 and f^{-1}(-17) = -4, since f(4) = 16 - 1 = 15 and f(-4) = -16 - 1 = -17. Check: \displaystyle \sqrt[3]{4 \cdot 16} = \sqrt[3]{64} = 4 and \displaystyle \sqrt[3]{4(-16)} = \sqrt[3]{-64} = -4.
- The points (-2, 9), (-1, 6), (0, 5), (1, 6), (2, 9) mirror to (9, -2), (6, -1), (5, 0), (6, 1), (9, 2). The input 6 now has the two outputs -1 and 1 (and 9 has -2 and 2), so the vertical line x = 6 meets the reflection twice: it is not a function.
- (a) and (c) only. y = x^3 and y = 2x - 1 are one to one. y = |x| and y = x^4 give the same output at -1 and 1, and y = 5 gives one output for every input; its reflection is the vertical line x = 5.
- (1, 1), (3, 3) and (-4, -4). Since u is increasing, solve u(x) = x: x^3 - 13x + 12 = 0, which factors as (x - 1)(x - 3)(x + 4) = 0.
- At (2, 2). v is increasing, so solve 3x - 4 = x, giving x = 2. Check with \displaystyle v^{-1}(x) = \frac{x + 4}{3}: v^{-1}(2) = 2.
- \displaystyle q^{-1}(x) = -\sqrt[3]{x}. q(1) = -1 and q^{-1}(1) = -1, so (1, -1) is on both; q(-1) = 1 and q^{-1}(-1) = 1, so (-1, 1) is on both; q(0) = q^{-1}(0) = 0. No contradiction: q is decreasing, and the result was for increasing functions.
- \displaystyle d\big(d(x)\big) = \frac{9}{9/x} = x for every x \ne 0, so d^{-1} = d. Mirror pairs on the graph: (1, 9) and (9, 1); (-1, -9) and (-9, -1). The graph is symmetric about y = x.
- The graph is a horizontal line y = c: every input gives the same output c, so the horizontal line y = c meets the graph at every point and the function is not one to one. Its reflection is the vertical line x = c, which is not the graph of a function.
- The line y = -x + c has slope -1, so it is perpendicular to y = x, and reflecting it in y = x maps it to itself (the point (a, c - a) goes to (c - a, a), and a = -(c - a) + c). Algebra: y = c - x solves to x = c - y, so the inverse is c - x again.
- Her inverse is right. Swapping coordinates gives the correct points whatever the scale. With unequal scales the line y = x is not at 45^\circ on the page and the swap is not a reflection of the drawing, so the curves do not look like mirror images.
- Symmetry about y = x means (b, a) is on the graph whenever (a, b) is. So f(a) = b gives f(b) = a, which says f^{-1}(b) = a = f(b) for every b in the range, and the graph of f^{-1}, the set of swapped points, is the graph of f itself. So f^{-1} = f.
- Take (p, q) and (r, s) on the line with p \ne r; then s \ne q because m \ne 0, and \displaystyle m = \frac{s - q}{r - p}. The mirror points (q, p) and (s, r) lie on the inverse, which has slope \displaystyle \frac{r - p}{s - q} = \frac{1}{m}.