1.4 Inverse functions (Page 3/11)

Calculus volume 1 Page 3 / 11

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 $[0, \infty) .$ The range of $f^{−1}$ is $[−2, \infty) .$ 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.

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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 $[0, \infty),$ we can define a new function $g$ such that the domain of $g$ is $[0, \infty)$ 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 $g^{−1} (x) = \sqrt{x} .$ On the other hand, the function $f (x) = x^{2}$ is also one-to-one on the domain $(− \infty, 0] .$ Therefore, we could also define a new function $h$ such that the domain of $h$ is $(− \infty, 0]$ 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 $h^{−1} (x) = − \sqrt{x}$ ( [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 $[0, \infty), g^{−1} (x) = \sqrt{x} .$ (b) For $h (x) = x^{2}$ restricted to $(− \infty, 0], h^{−1} (x) = − \sqrt{x} .$

Restricting the domain

Consider the function $f (x) = {(x + 1)}^{2} .$

Sketch the graph of $f$ and use the horizontal line test to show that $f$ is not one-to-one.
Show that $f$ is one-to-one on the restricted domain $[−1, \infty) .$ Determine the domain and range for the inverse of $f$ on this restricted domain and find a formula for $f^{−1} .$

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.
On the interval $[−1, \infty), f$ is one-to-one.

The domain and range of $f^{−1}$ are given by the range and domain of $f,$ respectively. Therefore, the domain of $f^{−1}$ is $[0, \infty)$ and the range of $f^{−1}$ is $[−1, \infty) .$ To find a formula for $f^{−1},$ solve the equation $y = {(x + 1)}^{2}$ for $x .$ If $y = {(x + 1)}^{2},$ then $x = −1 \pm \sqrt{y} .$ Since we are restricting the domain to the interval where $x \geq −1,$ we need $\pm \sqrt{y} \geq 0 .$ Therefore, $x = −1 + \sqrt{y} .$ Interchanging $x$ and $y,$ we write $y = −1 + \sqrt{x}$ and conclude that $f^{−1} (x) = −1 + \sqrt{x} .$

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Consider $f (x) = 1 / x^{2}$ restricted to the domain $(− \infty, 0) .$ 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 $(0, \infty) .$ The range of $f^{−1}$ is $(− \infty, 0) .$ The inverse function is given by the formula $f^{−1} (x) = −1 / \sqrt{x} .$

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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 $[- \frac{π}{2}, \frac{π}{2}] .$ 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 $[- \frac{π}{2}, \frac{π}{2}]$ 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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