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1.4 Inverse functions  (Page 3/11)

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An image of two graphs. The first graph is of “y = f(x)”, which is a curved increasing function, that increases at a faster rate as x increases. The point (a, b) is on the graph of the function in the first quadrant. The second graph also graphs “y = f(x)” with the point (a, b), but also graphs the function “y = f inverse (x)”, an increasing curved function, that increases at a slower rate as x increases. This function includes the point (b, a). In addition to the two functions, there is a diagonal dotted line potted with the equation “y =x”, which shows that “f(x)” and “f inverse (x)” are mirror images about the line “y =x”.
(a) The graph of this function f shows point ( a , b ) on the graph of f . (b) Since ( a , b ) is on the graph of f , the point ( b , a ) is on the graph of f −1 . The graph of f −1 is a reflection of the graph of f about the line y = x .

Sketching graphs of inverse functions

For the graph of f in the following image, sketch a graph of f −1 by sketching the line y = x and using symmetry. Identify the domain and range of f −1 .

An image of a graph. The x axis runs from -2 to 2 and the y axis runs from 0 to 2. The graph is of the function “f(x) = square root of (x +2)”, an increasing curved function. The function starts at the point (-2, 0). The x intercept is at (-2, 0) and the y intercept is at the approximate point (0, 1.4).

Reflect the graph about the line y = x . The domain of f −1 is [ 0 , ∞ ) . The range of f −1 is [ −2 , ∞ ) . By using the preceding strategy for finding inverse functions, we can verify that the inverse function is f −1 ( x ) = x 2 − 2 , as shown in the graph.

An image of a graph. The x axis runs from -2 to 2 and the y axis runs from -2 to 2. The graph is of two functions. The first function is “f(x) = square root of (x +2)”, an increasing curved function. The function starts at the point (-2, 0). The x intercept is at (-2, 0) and the y intercept is at the approximate point (0, 1.4). The second function is “f inverse (x) = (x squared) -2”, an increasing curved function that starts at the point (0, -2). The x intercept is at the approximate point (1.4, 0) and the y intercept is at the point (0, -2). In addition to the two functions, there is a diagonal dotted line potted with the equation “y =x”, which shows that “f(x)” and “f inverse (x)” are mirror images about the line “y =x”.
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Sketch the graph of f ( x ) = 2 x + 3 and the graph of its inverse using the symmetry property of inverse functions.


An image of a graph. The x axis runs from -3 to 4 and the y axis runs from -3 to 5. The graph is of two functions. The first function is “f(x) = 2x +3”, an increasing straight line function. The function has an x intercept at (-1.5, 0) and a y intercept at (0, 3). The second function is “f inverse (x) = (x - 3)/2”, an increasing straight line function, which increases at a slower rate than the first function. The function has an x intercept at (3, 0) and a y intercept at (0, -1.5). In addition to the two functions, there is a diagonal dotted line potted with the equation “y =x”, which shows that “f(x)” and “f inverse (x)” are mirror images about the line “y =x”.

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Restricting domains

As we have seen, f ( x ) = x 2 does not have an inverse function because it is not one-to-one. However, we can choose a subset of the domain of f such that the function is one-to-one. This subset is called a restricted domain    . By restricting the domain of f , we can define a new function g such that the domain of g is the restricted domain of f and g ( x ) = f ( x ) for all x in the domain of g . Then we can define an inverse function for g on that domain. For example, since f ( x ) = x 2 is one-to-one on the interval [ 0 , ∞ ) , we can define a new function g such that the domain of g is [ 0 , ∞ ) and g ( x ) = x 2 for all x in its domain. Since g is a one-to-one function, it has an inverse function, given by the formula g −1 ( x ) = x . On the other hand, the function f ( x ) = x 2 is also one-to-one on the domain ( − ∞ , 0 ] . Therefore, we could also define a new function h such that the domain of h is ( − ∞ , 0 ] and h ( x ) = x 2 for all x in the domain of h . Then h is a one-to-one function and must also have an inverse. Its inverse is given by the formula h −1 ( x ) = − x ( [link] ).

An image of two graphs. Both graphs have an x axis that runs from -2 to 5 and a y axis that runs from -2 to 5. The first graph is of two functions. The first function is “g(x) = x squared”, an increasing curved function that starts at the point (0, 0). This function increases at a faster rate for larger values of x. The second function is “g inverse (x) = square root of x”, an increasing curved function that starts at the point (0, 0). This function increases at a slower rate for larger values of x. The first function is “h(x) = x squared”, a decreasing curved function that ends at the point (0, 0). This function decreases at a slower rate for larger values of x. The second function is “h inverse (x) = -(square root of x)”, an increasing curved function that starts at the point (0, 0). This function decreases at a slower rate for larger values of x. In addition to the two functions, there is a diagonal dotted line potted with the equation “y =x”, which shows that “f(x)” and “f inverse (x)” are mirror images about the line “y =x”.
(a) For g ( x ) = x 2 restricted to [ 0 , ∞ ) , g −1 ( x ) = x . (b) For h ( x ) = x 2 restricted to ( − ∞ , 0 ] , h −1 ( x ) = − x .

Restricting the domain

Consider the function f ( x ) = ( x + 1 ) 2 .

  1. Sketch the graph of f and use the horizontal line test to show that f is not one-to-one.
  2. Show that f is one-to-one on the restricted domain [ −1 , ∞ ) . Determine the domain and range for the inverse of f on this restricted domain and find a formula for f −1 .
  1. The graph of f is the graph of y = x 2 shifted left 1 unit. Since there exists a horizontal line intersecting the graph more than once, f is not one-to-one.
    An image of a graph. The x axis runs from -6 to 6 and the y axis runs from -2 to 10. The graph is of the function “f(x) = (x+ 1) squared”, which is a parabola. The function decreases until the point (-1, 0), where it begins it increases. The x intercept is at the point (-1, 0) and the y intercept is at the point (0, 1). There is also a horizontal dotted line plotted on the graph, which crosses through the function at two points.
  2. On the interval [ −1 , ∞ ) , f is one-to-one.
    An image of a graph. The x axis runs from -6 to 6 and the y axis runs from -2 to 10. The graph is of the function “f(x) = (x+ 1) squared”, on the interval [1, infinity). The function starts from the point (-1, 0) and increases. The x intercept is at the point (-1, 0) and the y intercept is at the point (0, 1).
    The domain and range of f −1 are given by the range and domain of f , respectively. Therefore, the domain of f −1 is [ 0 , ∞ ) and the range of f −1 is [ −1 , ∞ ) . To find a formula for f −1 , solve the equation y = ( x + 1 ) 2 for x . If y = ( x + 1 ) 2 , then x = −1 ± y . Since we are restricting the domain to the interval where x ≥ −1 , we need ± y ≥ 0 . Therefore, x = −1 + y . Interchanging x and y , we write y = −1 + x and conclude that f −1 ( x ) = −1 + x .
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Consider f ( x ) = 1 / x 2 restricted to the domain ( − ∞ , 0 ) . Verify that f is one-to-one on this domain. Determine the domain and range of the inverse of f and find a formula for f −1 .

The domain of f −1 is ( 0 , ∞ ) . The range of f −1 is ( − ∞ , 0 ) . The inverse function is given by the formula f −1 ( x ) = −1 / x .

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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 [ − π 2 , π 2 ] . By doing so, we define the inverse sine function on the domain [ −1 , 1 ] such that for any x in the interval [ −1 , 1 ] , the inverse sine function tells us which angle θ in the interval [ − π 2 , π 2 ] satisfies sin θ = 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.
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Source:  OpenStax, Calculus volume 1. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11964/1.2
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