Lesson

Some general ideas about inverse functions have been presented in an earlier chapter.

While a *function *maps each element $x$`x` in its domain to an element $y$`y` in its range, the inverse function reverses the process by mapping each $y$`y` back to the original $x$`x`.

However, this will only work if each $y$`y` in the range is the image of exactly one $x$`x` from the domain. In other words, the function has to be a one-to-one function in order to have an inverse.

Functions that do have inverses are those that are something like a linear function, always increasing or always decreasing. In this situation, each element of the range is the image of exactly one domain element and so, the inverse mapping is possible. Functions of this kind are called *one-to-one* functions.

We can often get around the problem of a function not being one-to-one and therefore having no inverse, by restricting the domain of the function to a region over which the function is either increasing or decreasing and hence, one-to-one in that region.

We do this in the case of the trigonometric functions.

The sine function is increasing between $-\frac{\pi}{2}$−π2 and $\frac{\pi}{2}$π2. When the domain is restricted to this interval, the inverse function exists. It is called *arcsin*. The domain of the inverse sine function is the range of the sine function, the interval $[-1,1]$[−1,1], and the range of the inverse sine function is the restricted domain of sine function, the interval $[-\frac{\pi}{2},\frac{\pi}{2}]$[−π2,π2].

The inverse sine function is often notated $\sin^{-1}$`s``i``n`−1. Its graph has the same shape as the graph of the sine function. Imagine we start with the graph of $y=\sin x$`y`=`s``i``n``x` and rotate it $90^\circ$90° anticlockwise. Next, we flip the curve around the $y$`y`-axis. Finally, we chop off the parts of the curve above the line $y=\frac{\pi}{2}$`y`=π2 and below the line $y=\frac{-\pi}{2}$`y`=−π2. This final curve is the graph of the function $y=\sin^{-1}\left(x\right)$`y`=`s``i``n`−1(`x`), shown below.

The cosine function is decreasing between $0$0 and $\pi$π. When the domain is restricted to this interval, the inverse function exists. It is called *arccos*. The domain of the inverse cosine function is the range of the cosine function, the interval $[-1,1]$[−1,1], and the range of the inverse cosine function is the restricted domain of cosine function, the interval $[0,\pi]$[0,π].

On calculators, the notation for the *arccos *function is $\cos^{-1}$`c``o``s`−1. The shape of the graph is, again, the same as the shape of a section of the cosine graph but with a rotation and a flip, or we can think about it as a reflection in the line $y=x$`y`=`x`.

The tangent function is strictly increasing between $-\frac{\pi}{2}$−π2 and $\frac{\pi}{2}$π2. When the domain is restricted to this open interval, the inverse function exists. It is called *arctan*. The domain of the inverse tangent function is the range of the tangent function, the real numbers. The range of the inverse tangent function is the restricted domain of the tangent function, the interval $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$(−π2,π2).

Just as the tangent function has vertical asymptotes at $-\frac{\pi}{2}$−π2 and $\frac{\pi}{2}$π2, the inverse tangent function has horizontal asymptotes at these values. In the diagram below, the graphs of both functions have been sketched on the same set of axes. The line $y=x$`y`=`x` has been included to help make it clear that the graph of an *inverse function* has the shape of the graph of the function reflected in this line.

The idea that a function maps each element in its domain to an element in its range, and the inverse function (if there is one) takes the image back to the original elements, can be expressed concisely by writing

$f\left(f^{-1}(x)\right)=x$`f`(`f`−1(`x`))=`x`

where $f$`f` is a one-to-one function with inverse $f^{-1}$`f`−1. The range of $f$`f` is the domain of $f^{-1}$`f`−1 and the range of $f^{-1}$`f`−1 is the domain of $f$`f`.

For example, $\arctan(\tan x)=x$`a``r``c``t``a``n`(`t``a``n``x`)=`x` and $\tan(\arctan x)=x$`t``a``n`(`a``r``c``t``a``n``x`)=`x` .

Given that the inverse trigonometric functions have graphs that are reflections in the line $y=x$`y`=`x` of parts of the original functions, we should be able to express the inverses in terms of the original functions with the domains and ranges exchanged. How might this be done?

We could begin with the inverse sine function, $f(x)=\arcsin x$`f`(`x`)=`a``r``c``s``i``n``x`. Its domain is the interval $-1\le y\le1$−1≤`y`≤1 and its range is $-\frac{\pi}{2}\le f(x)\le\frac{\pi}{2}$−π2≤`f`(`x`)≤π2. If we now apply the sine function to $f(x)$`f`(`x`), which we can do because the domain of the sine function is the same as the range of $f$`f`, we have $\sin(f(x))=\sin(\arcsin x)=x$`s``i``n`(`f`(`x`))=`s``i``n`(`a``r``c``s``i``n``x`)=`x`, since sine and arcsine are inverses of one another. We can put $y=f(x)$`y`=`f`(`x`).

Evidently, $y=\arcsin x$`y`=`a``r``c``s``i``n``x` and $x=\sin y$`x`=`s``i``n``y` are equivalent statements if the domains and ranges match. The graph of $\arcsin x$`a``r``c``s``i``n``x` will look the same as the graph of $\sin y$`s``i``n``y`.

The point $\left(\frac{\pi}{3},\frac{1}{2}\right)$(π3,12) lies on the graph of $y=\cos x$`y`=`c``o``s``x`. What is the corresponding coordinate pair on the graph of $y=\cos^{-1}x$`y`=`c``o``s`−1`x`?

Consider the inverse cosine function, $y=\cos^{-1}x$`y`=`c``o``s`−1`x`.

What is its domain?

$\left(0,\pi\right)$(0,π)

A$\left[-1,1\right]$[−1,1]

B$\left(-1,1\right)$(−1,1)

C$\left(-\infty,\infty\right)$(−∞,∞)

D$\left[0,\pi\right]$[0,π]

E$\left(0,\pi\right)$(0,π)

A$\left[-1,1\right]$[−1,1]

B$\left(-1,1\right)$(−1,1)

C$\left(-\infty,\infty\right)$(−∞,∞)

D$\left[0,\pi\right]$[0,π]

EWhat is its range?

$\left[0,\pi\right]$[0,π]

A$\left(-\infty,\infty\right)$(−∞,∞)

B$\left[-1,1\right]$[−1,1]

C$\left(-1,1\right)$(−1,1)

D$\left(0,\pi\right)$(0,π)

E$\left[0,\pi\right]$[0,π]

A$\left(-\infty,\infty\right)$(−∞,∞)

B$\left[-1,1\right]$[−1,1]

C$\left(-1,1\right)$(−1,1)

D$\left(0,\pi\right)$(0,π)

EIs this function increasing or decreasing?

Decreasing

AIncreasing

BDecreasing

AIncreasing

BWhy is $\cos^{-1}x$

`c``o``s`−1`x`not defined for $x=-2$`x`=−2?$-2$−2 is not in the domain of $\cos^{-1}x$

`c``o``s`−1`x`.A$-2$−2 is not in the range of $\cos^{-1}x$

`c``o``s`−1`x`.B$\cos^{-1}x$

`c``o``s`−1`x`has an asymptote at $x=-2$`x`=−2.CYou cannot put negative numbers into $\cos^{-1}x$

`c``o``s`−1`x`.D$-2$−2 is not in the domain of $\cos^{-1}x$

`c``o``s`−1`x`.A$-2$−2 is not in the range of $\cos^{-1}x$

`c``o``s`−1`x`.B$\cos^{-1}x$

`c``o``s`−1`x`has an asymptote at $x=-2$`x`=−2.CYou cannot put negative numbers into $\cos^{-1}x$

`c``o``s`−1`x`.D

Consider the inverse tangent function, $y=\tan^{-1}x$`y`=`t``a``n`−1`x`.

What is its domain?

$\left(-\infty,\infty\right)$(−∞,∞)

A$\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$(−π2,π2)

B$\left(-1,1\right)$(−1,1)

C$\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$[−π2,π2]

D$\left[-1,1\right]$[−1,1]

E$\left(-\infty,\infty\right)$(−∞,∞)

A$\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$(−π2,π2)

B$\left(-1,1\right)$(−1,1)

C$\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$[−π2,π2]

D$\left[-1,1\right]$[−1,1]

EWhat is its range?

$\left[-1,1\right]$[−1,1]

A$\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$[−π2,π2]

B$\left(-\infty,\infty\right)$(−∞,∞)

C$\left(-1,1\right)$(−1,1)

D$\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$(−π2,π2)

E$\left[-1,1\right]$[−1,1]

A$\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$[−π2,π2]

B$\left(-\infty,\infty\right)$(−∞,∞)

C$\left(-1,1\right)$(−1,1)

D$\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$(−π2,π2)

EIs this function increasing or decreasing?

Increasing

ADecreasing

BIncreasing

ADecreasing

BWhich of the following value(s) of $x$

`x`is $y=\tan^{-1}x$`y`=`t``a``n`−1`x`not defined for?$x=\frac{\pi}{2}$

`x`=π2.ANone. $y=\tan^{-1}x$

`y`=`t``a``n`−1`x`is defined for all real $x$`x`.B$x=\frac{\pi}{2}$

`x`=π2 and $x=-\frac{\pi}{2}$`x`=−π2.C$x=0$

`x`=0.D$x=\frac{\pi}{2}$

`x`=π2.ANone. $y=\tan^{-1}x$

`y`=`t``a``n`−1`x`is defined for all real $x$`x`.B$x=\frac{\pi}{2}$

`x`=π2 and $x=-\frac{\pi}{2}$`x`=−π2.C$x=0$

`x`=0.D

Display and interpret the graphs of functions with the graphs of their inverse and/or reciprocal functions