7 Inverse Functions and L’Hôpital’s Rule

7.1 Inverse Functions

We say that two functions f and g are inverses if g(f(x))=x for all x in the domain of f and f(g(x))=x for all x in the domain of g. A function can only have an inverse if it is one-to-one, i.e. if we never have f(x1)=f(x2) for different elements x1 and x2 of the domain. This is equivalent to saying that the graph of the function passes the horizontal line test. The inverse of f is denoted f-1, which should not be confused with the function 1/f(x).

Key Idea 7.1.1 Inverse Functions

For a one-to–one function f,

  • The domain of f-1 is the range of f; the range of f-1 is the domain of f.

  • f-1(f(x))=x for all x in the domain of f.

  • f(f-1(x))=x for all x in the domain of f-1.

  • The graph of y=f-1(x) is the reflection across y=x of the graph of y=f(x).

  • y=f-1(x) if and only if f(y)=x and y is in the domain of f.

To determine whether or not f and g are inverses for each other, we check to see whether or not g(f(x))=x for all x in the domain of f,and f(g(x))=x for all x in the domain of g.

margin: -112-11(-0.5,0.375)(0.375,-0.5)(1,1.5)(1.5,1)xy Figure 7.1.1: A function f along with its inverse f-1. (Note how it does not matter which function we refer to as f; the other is f-1.) Λ
Example 7.1.1 Verifying Inverses

Determine whether or not the following pairs of functions are inverses:

  1. (a)

    f(x)=3x+1; g(x)=x-13

  2. (b)

    f(x)=x3+1; g(x)=x1/3-1


  1. (a)

    To check the composition we plug f(x) in for x in the definition of g as follows:


    So g(f(x))=x for all x in the domain of f. Likewise, you can check that f(g(x))=x for all x in the domain of g, so f and g are inverses.

  2. (b)

    If we try to proceed as before, we find that:


    This doesn’t seem to be the same as the identity function x. To verify this, we find a number a in the domain of f and show that g(f(a))a for that value. Let’s try x=1. Since f(1)=13+1=2, we find that g(f(1))=g(2)=21/3-10.26. Since g(f(1))1, these functions are not inverses.

Functions that are not one-to-one.

margin: -2224(-2,4)(2,4)xy Figure 7.1.2: The function f(x)=x2 is not one-to-one. Λ

Unfortunately, not every function we would like to find an inverse for is one-to-one. For example, the function f(x)=x2 is not one-to-one because f(-2)=f(2)=4. If f-1 is an inverse for f, then f-1(f(-2))=-2 implies that f-1(4)=-2. On the other hand, f-1(f(2))=2, so f-1(4)=2. We cannot have it both ways if f-1 is a function, so no such inverse exists. We can find a partial solution to this dilemma by restricting the domain of f. There are many possible choices, but traditionally we restrict the domain to the interval [0,). The function f-1(x)=x is now an inverse for this restricted version of f.

The inverse sine function

We consider the function f(x)=sinx, which is not one-to-one. A piece of the graph of f is in Figure 7.1.3(a). In order to find an appropriate restriction of the domain of f, we look for consecutive critical points where f takes on its minimum and maximum values. In this case, we use the interval [-π/2,π/2]. We define the inverse of f on this restricted range by y=sin-1x if and only if siny=x and -π/2yπ/2. The graph is a reflection of the graph of g across the line y=x, as seen in Figure 7.1.3(b).

-π-π2π2π-11xy -π2-11π2-π2-11π2sinxsin-1xxy (a) (b) Figure 7.1.3: (a) A portion of y=sinx. (b) A one-to-one portion of y=sinx along with y=sin-1x.

The inverse tangent function

Next we consider the function f(x)=tanx, which is also not one-to-one. A piece of the graph of f is given in Figure 7.1.4(a). In order to find an interval on which the function is one-to-one and on which the function takes on all values in the range, we use an interval between consecutive vertical asymptotes. Traditionally, the interval (-π/2,π/2) is chosen. Note that we choose the open interval in this case because the function f is not defined at the endpoints. So we define y=tan-1x if and only if tany=x and -π/2<y<π/2. The graph of y=tan-1x is shown in Figure 7.1.4(b). Also note that the vertical asymptotes of the original function are reflected to become horizontal asymptotes of the inverse function.

-3π2-π-π2π2π3π2-22xy -π2π2-π2π2tanxtan-1xxy (a) (b) Figure 7.1.4: (a) A portion of y=tanx. (b) A one-to-one portion of y=tanx along with y=tan-1.

The other inverse trigonometric functions are defined in a similar fashion. The resulting domains and ranges are summarized in Figure 7.1.5.

Function Restricted Domain Range Inverse Function Domain Range sinx [-π/2,π/2] [-1,1] sin-1x [-1,1] [-π/2,π/2] cosx [0,π] [-1,1] cos-1x [-1,1] [0,π] tanx (-π/2,π/2) (-,) tan-1x (-,) (-π/2,π/2) cscx [-π/2,0)(0,π/2] (-,-1][1,) csc-1x (-,-1][1,) [-π/2,0)(0,π/2] secx [0,π/2)(π/2,π] (-,-1][1,) sec-1x (-,-1][1,) [0,π/2)(π/2,π] cotx (0,π) (-,) cot-1x (-,) (0,π) Figure 7.1.5: Domains and ranges of the trigonometric and inverse trigonometric functions.
Example 7.1.2 Evaluating Inverse Trigonometric Functions

Find exact values for the following: margin: Sometimes, arcsin is used instead of sin-1. Similar “arc” functions are used for the other inverse trigonometric functions as well. Λ
(a) tan-1(1) (b) cos(sin-1(3/2)) (c) sin-1(sin(7π/6)) (d) tan(cos-1(11/15))


  1. (a)


  2. (b)


  3. (c)

    Since 7π/6 is not in the range of the inverse sine function, we should be careful with this one.

  4. (d)

    Since we don’t know the value of cos-1(11/15), we let θ stand for this value. We know that θ is an angle between 0 and π and that cos(θ)=11/15. In Figure 7.1.6, we use this information to construct a right triangle with angle θ, where the adjacent side over the hypotenuse must equal 11/15. Applying the Pythagorean Theorem we find that margin: θ11y15 Figure 7.1.6: A right triangle for the situation in Example 7.1.2 (4). Λ


    Finally, we have:


Exercises 7.1


Terms and Concepts

  1. 1.

    T/F: Every function has an inverse.

  2. 2.

    In your own words explain what it means for a function to be “one to one.”

  3. 3.

    If (1,10) lies on the graph of y=f(x), what can be said about the graph of y=f-1(x)?

  4. 4.

    If a function doesn’t have an inverse, what can we do to help it have an inverse?


In Exercises 5–6., given the graph of f, sketch the graph of f-1.

  1. 5.
  2. 6.

In Exercises 7–10., verify that the given functions are inverses.

  1. 7.

    f(x)=2x+6 and g(x)=12x-3

  2. 8.

    f(x)=x2+6x+11, x-3 and g(x)=x-2-3, x2

  3. 9.

    f(x)=3x-5, x5 and g(x)=3+5xx, x0

  4. 10.

    f(x)=x+1x-1, x1 and g(x)=f(x)

In Exercises 11–14., find a restriction of the domain of the given function on which the function will have an inverse.

  1. 11.


  2. 12.


  3. 13.


  4. 14.


In Exercises 15–18., find the inverse of the given function.

  1. 15.


  2. 16.


  3. 17.


  4. 18.


In Exercises 19–28., find the exact value.

  1. 19.


  2. 20.


  3. 21.


  4. 22.


  5. 23.


  6. 24.


  7. 25.


  8. 26.


  9. 27.


  10. 28.


In Exercises 29–32., simplify the expression.

  1. 29.


  2. 30.


  3. 31.


  4. 32.


  1. 33.

    Show that for any x in the domain of sec-1 we have sec-1x=cos-11x.

  2. 34.
    Show that for |x|1 we have cos-1x=π2-sin-1x. Hint: Recall the cofunction identity cosθ=sin(π2-θ) for all θ.
  3. 35.

    Show that for any x we have cot-1x=π2-tan-1x.

  4. 36.

    Show that for |x|1 we have csc-1x=π2-sec-1x.

  5. 37.

    A mass attached to a spring oscillates vertically about the equilibrium position y=0 according to the function y(t)=e-t(cos(3t)+13sin(3t)). Find the first positive time t for which y(t)=0.

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