Math out of the box (HUGH TESTING)
An inverse function reverses the operation done by a particular function. In other words, whatever a function does, the inverse function undoes it. In this section, we define an inverse function formally and state the necessary conditions for an inverse function to exist. We examine how to find an inverse function and study the relationship between the graph of a function and the graph of its inverse. Then we apply these ideas to define and discuss properties of the inverse trigonometric functions.
Existence of an Inverse Function
We begin with an example. Given a function ff and an output y=f(x),y=f(x), we are often interested in finding what value or values xx were mapped to yy by f.f. For example, consider the function f(x)=x3+4.f(x)=x3+4. Since any output y=x3+4,y=x3+4, we can solve this equation for xxto find that the input is x=y−4‾‾‾‾‾√3.x=y−43. This equation defines xx as a function of y.y. Denoting this function as f−1,f−1, and writing x=f−1(y)=y−4‾‾‾‾‾√3,x=f−1(y)=y−43, we see that for any xx in the domain of f,f−1(f(x))=f−1(x3+4)=x.f,f−1(f(x))=f−1(x3+4)=x. Thus, this new function, f−1,f−1,“undid” what the original function ff did. A function with this property is called the inverse function of the original function.
Given a function ff with domain DD and range R,R, its inverse function (if it exists) is the function f−1f−1 with domain RR and range DD such that f−1(y)=xf−1(y)=x if f(x)=y.f(x)=y. In other words, for a function ff and its inverse f−1,f−1,
Note that f−1f−1 is read as “f inverse.” Here, the −1−1 is not used as an exponent and f−1(x)≠1/f(x).f−1(x)≠1/f(x). Figure shows the relationship between the domain and range of f and the domain and range of f−1.f−1.
Recall that a function has exactly one output for each input. Therefore, to define an inverse function, we need to map each input to exactly one output. For example, let’s try to find the inverse function for f(x)=x2.f(x)=x2. Solving the equation y=x2y=x2 for x,x, we arrive at the equation x=±y√.x=±y. This equation does not describe xx as a function of yy because there are two solutions to this equation for every y>0.y>0. The problem with trying to find an inverse function for f(x)=x2f(x)=x2 is that two inputs are sent to the same output for each output y>0.y>0. The function f(x)=x3+4f(x)=x3+4 discussed earlier did not have this problem. For that function, each input was sent to a different output. A function that sends each input to a different output is called a one-to-one function.
We say a ff is a one-to-one function if f(x1)≠f(x2)f(x1)≠f(x2) when x1≠x2.x1≠x2.
One way to determine whether a function is one-to-one is by looking at its graph. If a function is one-to-one, then no two inputs can be sent to the same output. Therefore, if we draw a horizontal line anywhere in the xyxy-plane, according to the horizontal line test, it cannot intersect the graph more than once. We note that the horizontal line test is different from the vertical line test. The vertical line test determines whether a graph is the graph of a function. The horizontal line test determines whether a function is one-to-one (Figure).
A function ff is one-to-one if and only if every horizontal line intersects the graph of ff no more than once.
For each of the following functions, use the horizontal line test to determine whether it is one-to-one.
Is the function ff graphed in the following image one-to-one?
Finding a Function’s Inverse
We can now consider one-to-one functions and show how to find their inverses. Recall that a function maps elements in the domain of ff to elements in the range of f.f. The inverse function maps each element from the range of ff back to its corresponding element from the domain of f.f. Therefore, to find the inverse function of a one-to-one function f,f, given any yy in the range of f,f, we need to determine which xx in the domain of ff satisfies f(x)=y.f(x)=y. Since ff is one-to-one, there is exactly one such value x.x. We can find that value xx by solving the equation f(x)=yf(x)=y for x.x. Doing so, we are able to write xx as a function of yy where the domain of this function is the range of ff and the range of this new function is the domain of f.f. Consequently, this function is the inverse of f,f, and we write x=f−1(y).x=f−1(y).Since we typically use the variable xx to denote the independent variable and yy to denote the dependent variable, we often interchange the roles of xx and y,y, and write y=f−1(x).y=f−1(x). Representing the inverse function in this way is also helpful later when we graph a function ff and its inverse f−1f−1 on the same axes.
- Solve the equation y=f(x)y=f(x) for x.x.
- Interchange the variables xx and yy and write y=f−1(x).y=f−1(x).
Find the inverse for the function f(x)=3x−4.f(x)=3x−4. State the domain and range of the inverse function. Verify that f−1(f(x))=x.f−1(f(x))=x.
Find the inverse of the function f(x)=3x/(x−2).f(x)=3x/(x−2). State the domain and range of the inverse function.
Use the Note for finding inverse functions.
Graphing Inverse Functions
Let’s consider the relationship between the graph of a function ff and the graph of its inverse. Consider the graph of ff shown in Figure and a point (a,b)(a,b) on the graph. Since b=f(a),b=f(a), then f−1(b)=a.f−1(b)=a. Therefore, when we graph f−1,f−1, the point (b,a)(b,a) is on the graph. As a result, the graph of f−1f−1 is a reflection of the graph of ff about the line y=x.y=x.
For the graph of ff in the following image, sketch a graph of f−1f−1 by sketching the line y=xy=x and using symmetry. Identify the domain and range of f−1.f−1.
Sketch the graph of f(x)=2x+3f(x)=2x+3 and the graph of its inverse using the symmetry property of inverse functions.
The graphs are symmetric about the line y=x.y=x.
Restricting Domains
As we have seen, f(x)=x2f(x)=x2 does not have an inverse function because it is not one-to-one. However, we can choose a subset of the domain of ff such that the function is one-to-one. This subset is called a restricted domain. By restricting the domain of f,f, we can define a new function gg such that the domain of gg is the restricted domain of ff and g(x)=f(x)g(x)=f(x) for all xx in the domain of g.g. Then we can define an inverse function for gg on that domain. For example, since f(x)=x2f(x)=x2 is one-to-one on the interval [0,∞),[0,∞), we can define a new function gg such that the domain of gg is [0,∞)[0,∞) and g(x)=x2g(x)=x2 for all xx in its domain. Since gg is a one-to-one function, it has an inverse function, given by the formula g−1(x)=x√.g−1(x)=x. On the other hand, the function f(x)=x2f(x)=x2 is also one-to-one on the domain (−∞,0].(−∞,0]. Therefore, we could also define a new function hh such that the domain of hh is (−∞,0](−∞,0] and h(x)=x2h(x)=x2 for all xx in the domain of h.h. Then hh is a one-to-one function and must also have an inverse. Its inverse is given by the formula h−1(x)=−x√h−1(x)=−x (Figure).
Consider the function f(x)=(x+1)2.f(x)=(x+1)2.
- Sketch the graph of ff and use the horizontal line test to show that ff is not one-to-one.
- Show that ff is one-to-one on the restricted domain [−1,∞).[−1,∞). Determine the domain and range for the inverse of ff on this restricted domain and find a formula for f−1.f−1.
Consider f(x)=1/x2f(x)=1/x2 restricted to the domain (−∞,0).(−∞,0). Verify that ff is one-to-one on this domain. Determine the domain and range of the inverse of ff and find a formula for f−1.f−1.
The domain and range of f−1f−1 is given by the range and domain of f,f, respectively. To find f−1,f−1, solve y=1/x2y=1/x2 for x.x.
Inverse Trigonometric Functions
The six basic trigonometric functions are periodic, and therefore they are not one-to-one. However, if we restrict the domain of a trigonometric function to an interval where it is one-to-one, we can define its inverse. Consider the sine function ([link]). The sine function is one-to-one on an infinite number of intervals, but the standard convention is to restrict the domain to the interval [−π2,π2].[−π2,π2].By doing so, we define the inverse sine function on the domain [−1,1][−1,1] such that for any xx in the interval [−1,1],[−1,1], the inverse sine function tells us which angle θθ in the interval [−π2,π2][−π2,π2] satisfies sinθ=x.sinθ=x. Similarly, we can restrict the domains of the other trigonometric functions to define inverse trigonometric functions, which are functions that tell us which angle in a certain interval has a specified trigonometric value.
The inverse sine function, denoted sin−1sin−1 or arcsin, and the inverse cosine function, denoted cos−1cos−1 or arccos, are defined on the domain D={x∣∣−1≤x≤1}D={x|−1≤x≤1} as follows:
The inverse tangent function, denoted tan−1tan−1 or arctan, and inverse cotangent function, denoted cot−1cot−1 or arccot, are defined on the domain D={x∣∣−∞<x<∞}D={x|−∞<x<∞} as follows:
The inverse cosecant function, denoted csc−1csc−1 or arccsc, and inverse secant function, denoted sec−1sec−1 or arcsec, are defined on the domain D={x∣∣|x|≥1}D={x||x|≥1} as follows:
To graph the inverse trigonometric functions, we use the graphs of the trigonometric functions restricted to the domains defined earlier and reflect the graphs about the line y=xy=x (Figure).
Go to the following site for more comparisons of functions and their inverses.
When evaluating an inverse trigonometric function, the output is an angle. For example, to evaluate cos−1(12),cos−1(12), we need to find an angle θθ such that cosθ=12.cosθ=12. Clearly, many angles have this property. However, given the definition of cos−1,cos−1, we need the angle θθ that not only solves this equation, but also lies in the interval [0,π].[0,π]. We conclude that cos−1(12)=π3.cos−1(12)=π3.
We now consider a composition of a trigonometric function and its inverse. For example, consider the two expressions sin(sin−1(2√2))sin(sin−1(22)) and sin−1(sin(π)).sin−1(sin(π)). For the first one, we simplify as follows:
For the second one, we have
The inverse function is supposed to “undo” the original function, so why isn’t sin−1(sin(π))=π?sin−1(sin(π))=π? Recalling our definition of inverse functions, a function ff and its inverse f−1f−1 satisfy the conditions f(f−1(y))=yf(f−1(y))=y for all yy in the domain of f−1f−1 and f−1(f(x))=xf−1(f(x))=x for all xx in the domain of f,f, so what happened here? The issue is that the inverse sine function, sin−1,sin−1, is the inverse of the restricted sine function defined on the domain [−π2,π2].[−π2,π2]. Therefore, for xx in the interval [−π2,π2],[−π2,π2], it is true that sin−1(sinx)=x.sin−1(sinx)=x. However, for values of xx outside this interval, the equation does not hold, even though sin−1(sinx)sin−1(sinx) is defined for all real numbers x.x.
What about sin(sin−1y)?sin(sin−1y)? Does that have a similar issue? The answer is no. Since the domain of sin−1sin−1 is the interval [−1,1],[−1,1], we conclude that sin(sin−1y)=ysin(sin−1y)=y if −1≤y≤1−1≤y≤1 and the expression is not defined for other values of y.y. To summarize,
and
Similarly, for the cosine function,
and
Similar properties hold for the other trigonometric functions and their inverses.
Evaluate each of the following expressions.
- sin−1(−3√2)sin−1(−32)
- tan(tan−1(−13√))tan(tan−1(−13))
- cos−1(cos(5π4))cos−1(cos(5π4))
- sin−1(cos(2π3))sin−1(cos(2π3))
In many areas of science, engineering, and mathematics, it is useful to know the maximum value a function can obtain, even if we don’t know its exact value at a given instant. For instance, if we have a function describing the strength of a roof beam, we would want to know the maximum weight the beam can support without breaking. If we have a function that describes the speed of a train, we would want to know its maximum speed before it jumps off the rails. Safe design often depends on knowing maximum values.
This project describes a simple example of a function with a maximum value that depends on two equation coefficients. We will see that maximum values can depend on several factors other than the independent variable x.
- Consider the graph in Figure of the function y=sinx+cosx.y=sinx+cosx. Describe its overall shape. Is it periodic? How do you know?
Using a graphing calculator or other graphing device, estimate the xx– and yy-values of the maximum point for the graph (the first such point where x > 0). It may be helpful to express the xx-value as a multiple of π.
- Now consider other graphs of the form y=Asinx+Bcosxy=Asinx+Bcosx for various values of A and B. Sketch the graph when A = 2 and B = 1, and find the xx– and y-values for the maximum point. (Remember to express the x-value as a multiple of π, if possible.) Has it moved?
- Repeat for A = 1, B = 2. Is there any relationship to what you found in part (2)?
- Complete the following table, adding a few choices of your own for A and B:
A B x y A B x y 0 1 3‾√3 1 1 0 1 3‾√3 1 1 12 5 1 2 5 12 2 1 2 2 3 4 4 3 - Try to figure out the formula for the y-values.
- The formula for the xx-values is a little harder. The most helpful points from the table are (1,1),(1,3‾√),(3‾√,1).(1,1),(1,3),(3,1). (Hint: Consider inverse trigonometric functions.)
- If you found formulas for parts (5) and (6), show that they work together. That is, substitute the xx-value formula you found into y=Asinx+Bcosxy=Asinx+Bcosx and simplify it to arrive at the yy-value formula you found.
Key Concepts
- For a function to have an inverse, the function must be one-to-one. Given the graph of a function, we can determine whether the function is one-to-one by using the horizontal line test.
- If a function is not one-to-one, we can restrict the domain to a smaller domain where the function is one-to-one and then define the inverse of the function on the smaller domain.
- For a function ff and its inverse f−1,f(f−1(x))=xf−1,f(f−1(x))=x for all xx in the domain of f−1f−1 and f−1(f(x))=xf−1(f(x))=x for all xx in the domain of f.f.
- Since the trigonometric functions are periodic, we need to restrict their domains to define the inverse trigonometric functions.
- The graph of a function ff and its inverse f−1f−1 are symmetric about the line y=x.y=x.
Key Equations
- Inverse functions
f−1(f(x))=xfor allxinD,andf(f−1(y))=yfor allyinR.f−1(f(x))=xfor allxinD,andf(f−1(y))=yfor allyinR.
For the following exercises, use the horizontal line test to determine whether each of the given graphs is one-to-one.
For the following exercises, a. find the inverse function, and b. find the domain and range of the inverse function.
f(x)=x2−4,x≥0f(x)=x2−4,x≥0
f(x)=x−4‾‾‾‾‾√3f(x)=x−43
f(x)=x3+1f(x)=x3+1
f(x)=(x−1)2,x≤1f(x)=(x−1)2,x≤1
f(x)=x−1‾‾‾‾‾√f(x)=x−1
f(x)=1x+2f(x)=1x+2
For the following exercises, use the graph of ff to sketch the graph of its inverse function.
For the following exercises, use composition to determine which pairs of functions are inverses.
f(x)=8x,g(x)=x8f(x)=8x,g(x)=x8
f(x)=8x+3,g(x)=x−38f(x)=8x+3,g(x)=x−38
f(x)=5x−7,g(x)=x+57f(x)=5x−7,g(x)=x+57
f(x)=23x+2,g(x)=32x+3f(x)=23x+2,g(x)=32x+3
f(x)=1x−1,x≠1,g(x)=1x+1,x≠0f(x)=1x−1,x≠1,g(x)=1x+1,x≠0
f(x)=x3+1,g(x)=(x−1)1/3f(x)=x3+1,g(x)=(x−1)1/3
f(x)=x2+2x+1,x≥−1,g(x)=−1+x√,x≥0f(x)=x2+2x+1,x≥−1,g(x)=−1+x,x≥0
f(x)=4−x2‾‾‾‾‾‾√,0≤x≤2,g(x)=4−x2‾‾‾‾‾‾√,0≤x≤2f(x)=4−x2,0≤x≤2,g(x)=4−x2,0≤x≤2
For the following exercises, evaluate the functions. Give the exact value.
tan−1(3√3)tan−1(33)
cos−1(−2√2)cos−1(−22)
cot−1(1)cot−1(1)
sin−1(−1)sin−1(−1)
cos−1(3√2)cos−1(32)
cos(tan−1(3‾√))cos(tan−1(3))
sin(cos−1(2√2))sin(cos−1(22))
sin−1(sin(π3))sin−1(sin(π3))
tan−1(tan(−π6))tan−1(tan(−π6))
The function C=T(F)=(5/9)(F−32)C=T(F)=(5/9)(F−32) converts degrees Fahrenheit to degrees Celsius.
- Find the inverse function F=T−1(C)F=T−1(C)
- What is the inverse function used for?
[T] The velocity V (in centimeters per second) of blood in an artery at a distance x cm from the center of the artery can be modeled by the function V=f(x)=500(0.04−x2)V=f(x)=500(0.04−x2) for 0≤x≤0.2.0≤x≤0.2.
- Find x=f−1(V).x=f−1(V).
- Interpret what the inverse function is used for.
- Find the distance from the center of an artery with a velocity of 15 cm/sec, 10 cm/sec, and 5 cm/sec.
A function that converts dress sizes in the United States to those in Europe is given by D(x)=2x+24.D(x)=2x+24.
- Find the European dress sizes that correspond to sizes 6, 8, 10, and 12 in the United States.
- Find the function that converts European dress sizes to U.S. dress sizes.
- Use part b. to find the dress sizes in the United States that correspond to 46, 52, 62, and 70.
[T] The cost to remove a toxin from a lake is modeled by the function
C(p)=75p/(85−p),C(p)=75p/(85−p), where CC is the cost (in thousands of dollars) and pp is the amount of toxin in a small lake (measured in parts per billion [ppb]). This model is valid only when the amount of toxin is less than 85 ppb.
- Find the cost to remove 25 ppb, 40 ppb, and 50 ppb of the toxin from the lake.
- Find the inverse function. c. Use part b. to determine how much of the toxin is removed for $50,000.
[T] A race car is accelerating at a velocity given by
v(t)=254t+54,v(t)=254t+54,
where v is the velocity (in feet per second) at time t.
- Find the velocity of the car at 10 sec.
- Find the inverse function.
- Use part b. to determine how long it takes for the car to reach a speed of 150 ft/sec.
[T] An airplane’s Mach number M is the ratio of its speed to the speed of sound. When a plane is flying at a constant altitude, then its Mach angle is given by μ=2sin−1(1M).μ=2sin−1(1M).
Find the Mach angle (to the nearest degree) for the following Mach numbers.
- μ=1.4μ=1.4
- μ=2.8μ=2.8
- μ=4.3μ=4.3
[T] Using μ=2sin−1(1M),μ=2sin−1(1M), find the Mach number M for the following angles.
- μ=π6μ=π6
- μ=2π7μ=2π7
- μ=3π8μ=3π8
[T] The temperature (in degrees Celsius) of a city in the northern United States can be modeled by the function
T(x)=5+18sin[π6(x−4.6)],T(x)=5+18sin[π6(x−4.6)],
where xx is time in months and x=1.00x=1.00 corresponds to January 1. Determine the month and day when the temperature is 21°C.21°C.
[T] The depth (in feet) of water at a dock changes with the rise and fall of tides. It is modeled by the function
D(t)=5sin(π6t−7π6)+8,D(t)=5sin(π6t−7π6)+8,
where tt is the number of hours after midnight. Determine the first time after midnight when the depth is 11.75 ft.
[T] An object moving in simple harmonic motion is modeled by the function
s(t)=−6cos(πt2),s(t)=−6cos(πt2),
where ss is measured in inches and tt is measured in seconds. Determine the first time when the distance moved is 4.5 ft.
[T] A local art gallery has a portrait 3 ft in height that is hung 2.5 ft above the eye level of an average person. The viewing angle θθcan be modeled by the function
θ=tan−15.5x−tan−12.5x,θ=tan−15.5x−tan−12.5x,
where xx is the distance (in feet) from the portrait. Find the viewing angle when a person is 4 ft from the portrait.
[T] Use a calculator to evaluate tan−1(tan(2.1))tan−1(tan(2.1)) and cos−1(cos(2.1)).cos−1(cos(2.1)). Explain the results of each.
[T] Use a calculator to evaluate sin(sin−1(−2))sin(sin−1(−2)) and tan(tan−1(−2)).tan(tan−1(−2)). Explain the results of each.
Glossary
- horizontal line test
- a function ff is one-to-one if and only if every horizontal line intersects the graph of f,f, at most, once
- inverse function
- for a function f,f, the inverse function f−1f−1 satisfies f−1(y)=xf−1(y)=x if f(x)=yf(x)=y
- inverse trigonometric functions
- the inverses of the trigonometric functions are defined on restricted domains where they are one-to-one functions
- one-to-one function
- a function ff is one-to-one if f(x1)≠f(x2)f(x1)≠f(x2) if x1≠x2x1≠x2
- restricted domain
- a subset of the domain of a function ff
Use the horizontal line test.