DITF - Inverse Functions Lesson

Inverse Functions

The word inverse has various meanings in mathematics, all of which relate to "undoing" or "reversing" an operation. The additive inverse of x is -x because their sum is 0, the additive identity, and the multiplicative inverse of x is 1/x because their product is 1, the multiplicative identity. Inverse, in this context, is another term for reciprocal.

Definition

The term inverse has yet another meaning when applied to functions. Two functions f and g are inverses if composing them, in any order, gives the identity function, x. Thus the inverse of a function f is another function, denoted by f ^-1, which undoes the effect of the original function. Do not mistake the -1 in f ^-1 for an exponent; f ^-1(x) does not mean 1/f(x). The reciprocal 1/f(x) would be written as [f(x)]^-1.

More formally, suppose f and g are functions. If f(g(x)) = x for each x in the domain of g and g(f(x)) = x for each x in the domain of f, then f and g are inverse functions. The notation for expressing this relationship is g = f ^-1and f = g^-1. Note that an inverse function is not the reciprocal function.

Existence of an Inverse Function

Every function expresses a correspondence (mapping) between its domain and its range; to an input x from the domain, f assigns f(x), exactly one member of the range. An inverse function, if one exists, reverses the correspondence. To be invertible, a term often used to describe functions possessing an inverse, f must be one-to-one, which means it never takes on the same value twice. Just like not every relation is a function, not every function has an inverse function. The following alternate definition relies on a function being one-to-one. Let f be a one-to-one function with domain A and range B. Then its inverse function f ^-1has domain B and range A and is defined by equation image indicator $LaTeX: f^{-1}\left(y\right)=x\Leftrightarrow f\left(x\right)=y$ $f^{-1}\left(y\right)=x\Leftrightarrow f\left(x\right)=y$ for any y in B. You can think of a one-to-one function as different x values not having the same y values. It may be helpful to associate a graphical representation of functions that are one-to-one and those that are not. The first graphed function below is one-to-one but the second graphed function is not one-to-one.

graph of one-to-one function graph of a non-one-to-one function

Horizontal Line Test for One-to-One Functions

Just as the vertical-line test is a quick way to visually determine if a graph represents a function, the horizontal line test is helpful in determining if a function is one-to-one.

Any horizontal line drawn on the first graph above intersects the curve only once, which means the function is one-to-one, i.e., whenever x₁≠x₂ then f(x₁) ≠ f(x₂).
Any horizontal line drawn on the second curve above intersects the curve more than once, which shows the function is not one-to-one, i.e., whenever x₁≠ x₂ then f(x₁) = f(x₂).

Graphs of Inverse Functions

The graphs of a function and its inverse are closely related, as illustrated in the animation shown below.

graph of inverse functions The graphs of f and f ^-1 appear to be mirror images of each other with respect to the line y = x, which means the graph of f ^-1 is obtained by reflecting the graph of f about the line y = x. The graph of f contains the point (a,b) if and only if the graph of $LaTeX: f^{-1}$ $f^{-1}$ contains the point (b,a). The domain of f ^-1 is the range of f, and the range of f ^-1 is the domain of f. This relationship is defined as equation image indicator $LaTeX: f^{-1}\left(y\right)=x\Leftrightarrow f\left(x\right)=y$ $f^{-1}\left(y\right)=x\Leftrightarrow f\left(x\right)=y$ . To maintain consistency in graphing functions and their inverses, the input values (independent variable) of f ^-1 lie along the x-axis, rather than along the y-axis as the previous statements might suggest.

Finding an Inverse Function

The inverse function of a one-to-one function is obtained as follows:

1. Solve the equation y = f(x) for x, which produces a formula x = f ^-1(y) where x is expressed as a function of y.

2. Interchange x and y, obtaining a formula y = f ^-1(x) where f ^-1 is expressed in the conventional format with x as the independent variable and y as the dependent variable.

3. Define the domain of f ^-1 to be the range of f.

4. Verify that f(f ^-1(x)) = x and f ^-1(f (x)) = x.

View the presentation on defining inverse functions, determining if a function has an inverse function, and finding the equation of an inverse function.

Logarithmic and Exponential Inverse Functions

One of the most important and useful inverse function pairs in calculus is the logarithmic-exponential pair. Logarithms are the foundation of logarithmic functions and are the inverses of exponents whenever the base of the logarithm is a positive constant. For a positive base a and real numbers x and y, logarithms and exponents are related as follows:

equation image indicator $LaTeX: y=\log_ax\Leftrightarrow x=a^y$ $y=\log_ax\Leftrightarrow x=a^y$

A function of the form f(x) = a^x, where a is a positive constant is an exponential function. Exponential functions are characterized by a variable in the exponent rather than a variable in the base, as in algebraic functions. View the presentation below that reviews how to solve exponential equations.

Both logarithmic and exponential functions are transcendental functions, i.e., they are not algebraic. The presentation below focuses on the graphs of general exponential and logarithmic functions and their corresponding characteristics.

A review of logarithms, converting between logarithmic and exponential forms, and graphing a logarithmic function are presented in the video below. View the presentation from the beginning to 7:57.

The most commonly used bases of a logarithm are 10, referred to as a common logarithm and written as log x, and e, referred to as a natural logarithm and written as ln x, although others may be used (e.g., log₂ x). The presentation below applies the natural logarithm function to exponential equation solving.

Derivative of an Inverse Function

Consider a function f that is one-to-one, continuous, and defined on an interval I. Recall that the graph of a continuous function has no break in it. Since the graph of f ^-1 is obtained by reflecting f about the line y = x, the graph of f ^-1 has no break in it either. Thus, it follows that the inverse function f ^-1 is also continuous.

Now consider a function f that is one-to-one and differentiable on an interval I. The graph of a differentiable function has no corner or kink in it. Following the same line of reasoning as before, the graph of f ^-1 has no corner or kink in it either. Therefore, it is reasonable to expect that f ^-1 is also differentiable, except where its tangents are vertical.

graph of line and its reciprocals By examining the graph of a line and its inverse, it is apparent that the slopes of f and f ^-1 are reciprocals of each other. It is always the case that reflecting any nonhorizontal or nonvertical line across the line y = x always inverts the line's slope. If the original line has slope m≠0, the reflected line has slope 1/m. This reciprocal relationship also holds for other functions, but care must be taken to compare slopes at corresponding points. If the slope of f at the point (a, f(a)) is f'(a)

and f'(a)≠0, then the slope of y = f ^-1(x) at the point (f(a), a) is the reciprocal 1/ f'(a). More formally, if f is a one-to-one differentiable function on an interval I with inverse function g = f ^-1 and f'(g(a)) ≠ 0, then the inverse function is differentiable at a and equation image indicator $LaTeX: g'\left(a\right)=\frac{1}{f'\left(g\left(a\right)\right)}$ $g'\left(a\right)=\frac{1}{f'\left(g\left(a\right)\right)}$ . This relationship may be written more generally as

equation image indicator $LaTeX: g'\left(x\right)=\frac{1}{f'\left(g\left(x\right)\right)},\:f'\left(g\left(x\right)\right)\ne0$ $g'\left(x\right)=\frac{1}{f'\left(g\left(x\right)\right)},\:f'\left(g\left(x\right)\right)\ne0$ . Examine how this theorem is applied in the example below.

Example

If equation image indicator $LaTeX: f\left(x\right)=2x+\cos x,\:find\left(f^{-1}\right)'\left(1\right)$ $f\left(x\right)=2x+\cos x,\:find\left(f^{-1}\right)'\left(1\right)$ .

f is one-to-one because f'(x) = 2 - sin x > 0 and always increasing.
f(0) = 1 and $LaTeX: \left(f^{-1}\right)\left(1\right)=0$ $\left(f^{-1}\right)\left(1\right)=0$
$LaTeX: \left(f^{-1}\right)^1\left(1\right)=\frac{1}{f'\left(f^{-1}\right)\left(1\right)}=\frac{1}{f'\left(0\right)}=\frac{1}{2-\sin0}=\frac{1}{2}$ $\left(f^{-1}\right)^1\left(1\right)=\frac{1}{f'\left(f^{-1}\right)\left(1\right)}=\frac{1}{f'\left(0\right)}=\frac{1}{2-\sin0}=\frac{1}{2}$

Confirmation:

At the point (0, 1), f '(0) = 2 - sin(0) = 2.
At the point (1, 0),
Slopes of inverse functions at corresponding points are reciprocals.

An alternative notation for the reciprocal relationship existing between the slope of a function and the slope of its inverse is often helpful.

If equation image indicator $LaTeX: y=g\left(x\right)=f^{-1}\left(x\right)$ $y=g\left(x\right)=f^{-1}\left(x\right)$ , then $LaTeX: f\left(y\right)=x$ $f\left(y\right)=x$ and $LaTeX: f'\left(y\right)=\frac{dx}{dy}$ $f'\left(y\right)=\frac{dx}{dy}$ .

equation image indicator $LaTeX: g'\left(x\right)=\frac{dy}{dx}=\frac{1}{f'\left(g\left(x\right)\right)}=\frac{1}{f'\left(y\right)}=\frac{1}{\frac{dx}{dy}}\\ \frac{dy}{dx}=\frac{1}{\frac{dx}{dy}}$ $g'\left(x\right)=\frac{dy}{dx}=\frac{1}{f'\left(g\left(x\right)\right)}=\frac{1}{f'\left(y\right)}=\frac{1}{\frac{dx}{dy}}\\ \frac{dy}{dx}=\frac{1}{\frac{dx}{dy}}$

Explore derivatives by CLICKING HERE Links to an external site.. The derivative widget provides self-check opportunities with a variety of self-entered problems. Simply enter the words derivative of followed by the expression, e.g., derivative of sin^-1(2x).

Inverse Functions Practice

Inverse Functions: Even More Problems!

Complete problems from your textbook and/or online resources as needed to ensure your complete understanding of Inverse Functions.

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