5.1 Inverse Functions

Vanessa Coffelt

5.1 Inverse Functions

In Section 1.2, we defined functions as processes. In this section, we seek to reverse, or `undo’ those processes. As in real life, we will find that some processes (like putting on socks and shoes) are reversible while some (like baking a cake) are not.

Consider the function $f(x) = 3x+4$ . Starting with a real number input $x$ , we apply two steps in the following sequence: first we multiply the input by $3$ and, second, we add $4$ to the result.

To reverse this process, we seek a function $g$ which will undo each of these steps and take the output from $f$ , $3x+4$ , and return the input $x$ . If we think of the two-step process of first putting on socks then putting on shoes, to reverse the process, we first take off the shoes and then we take off the socks. In much the same way, the function $g$ should undo each step of $f$ but in the opposite order. That is, the function $g$ should first subtract $4$ from the input $x$ then divide the result by $3$ . This leads us to the formula $g(x) = \frac{x-4}{3}$ .

Let’s check to see if the function $g$ does the job. If $x=5$ , then $f(5) = 3(5)+4 = 15+4 = 19$ . Taking the output 19 from $f$ , we substitute it into $g$ to get $g(19) = \frac{19-4}{3} = \frac{15}{3} = 5$ , which is our original input to $f$ . To check that $g$ does the job for all $x$ in the domain of $f$ , we take the generic output from $f$ , $f(x) = 3x+4$ , and substitute that into $g$ . That is, we simplify $g(f(x)) = g(3x+4) = \frac{(3x+4)-4}{3} = \frac{3x}{3} = x$ , which is our original input to $f$ . If we carefully examine the arithmetic as we simplify $g(f(x))$ , we actually see $g$ first `undoing’ the addition of $4$ , and then `undoing’ the multiplication by $3$ .

Not only does $g$ undo $f$ , but $f$ also undoes $g$ . That is, if we take the output from $g$ , $g(x) = \frac{x-4}{3}$ , and substitute that into $f$ , we get $f(g(x)) = f\left(\frac{x-4}{3}\right) = 3 \left(\frac{x-4}{3}\right) + 4 = (x-4) + 4 = x$ . Using the language of function composition developed in Section 1.5.2, the statements $g(f(x)) = x$ and $f(g(x)) = x$ can be written as $(g \circ f)(x) = x$ and $(f \circ g)(x) = x$ , respectively.^[1] Abstractly, we can visualize the relationship between $f$ and $g$ in the diagram below.

Two ovals, the first representing the domain of the function f and the second representing the range of the function f. There is a point x = g(f(x)) labeled in the first oval and y = f(x) is a point in the second oval. There is an arrow drawn from the first oval to the second labeled f. A second arrow goes from the second oval to the first, the path is dashed, and is labeled g. — Visual Representation between f and g

The main idea to get from the diagram is that $g$ takes the outputs from $f$ and returns them to their respective inputs, and conversely, $f$ takes outputs from $g$ and returns them to their respective inputs. We now have enough background to state the central definition of the section.

Definition 5.1

Suppose $f$ and $g$ are two functions such that

$(g \circ f)(x) = x$ for all $x$ in the domain of $f$ and
$(f \circ g)(x) = x$ for all $x$ in the domain of $g$

then $f$ and $g$ are inverses of each other and the functions $f$ and $g$ are said to be invertible.

If we abstract one step further, we can express the sentiment in Definition 5.1 by saying that $f$ and $g$ are inverses if and only if $g \circ f = I_{1}$ and $f \circ g = I_{2}$ where $I_{1}$ is the identity function restricted^[2] to the domain of $f$ and $I_{2}$ is the identity function restricted to the domain of $g$ .

In other words, $I_{1}(x) = x$ for all $x$ in the domain of $f$ and $I_{2}(x) = x$ for all $x$ in the domain of $g$ . Using this description of inverses along with the properties of function composition listed in Theorem 1.6, we can show that function inverses are unique.^[3]

Suppose $g$ and $h$ are both inverses of a function $f$ . By Theorem 5.1, the domain of $g$ is equal to the domain of $h$ , because both are the range of $f$ . This means the identity function $I_{2}$ applies both to the domain of $h$ and the domain of $g.$ Thus

$\begin{array}{rcl} h &=& h \circ I_{2} \\ &=& h \circ (f \circ g) \\ &=& (h \circ f) \circ g \\ &=& I_{1} \circ g \\ &=& g \end{array}$

as required.

We summarize the important properties of invertible functions in the following theorem. Apart from introducing notation, each of the results below are immediate consequences of the idea that inverse functions map the outputs from a function $f$ back to their corresponding inputs.

Theorem 5.1 Properties of Inverse Functions

Suppose $f$ is an invertible function.

There is exactly one inverse function for $f$ , denoted $f^{-1}$ (read ` $f$ -inverse’)
The range of $f$ is the domain of $f^{-1}$ and the domain of $f$ is the range of $f^{-1}$
$f(a) = c$ if and only if $a = f^{-1}(c)$
NOTE: In particular, for all $y$ in the range of $f$ , the solution to $f(x) = y$ is $x = f^{-1}(y)$
$(a,c)$ is on the graph of $f$ if and only if $(c,a)$ is on the graph of $f^{-1}$
NOTE: This means graph of $y=f^{-1}(x)$ is the reflection of the graph of $y=f(x)$ across $y=x$ .^[4]
$f^{-1}$ is an invertible function and $(f^{-1})^{-1} = f$

The notation $f^{-1}$ is an unfortunate choice because you’ve been programmed since Algebra I to think of this as $\frac{1}{f}$ . This is most definitely not the case, for instance, $f(x) = 3x+4$ has as its inverse $f^{-1}(x) = \frac{x-4}{3}$ , which is certainly different than $\frac{1}{f(x)} = \frac{1}{3x+4}.$

Why does this confusing notation persist? As we mentioned in Section 1.5.2, the identity function $I$ is to function composition what the real number $1$ is to real number multiplication. The choice of notation $f^{-1}$ alludes to the property that $f^{-1} \circ f = I_{1}$ and $f \circ f^{-1} = I_{2}$ , in much the same way as $3^{-1} \cdot 3 = 1$ and $3 \cdot 3^{-1} = 1$ .

Before we embark on an example, we demonstrate the pertinent parts of Theorem 5.1 to the inverse pair $f(x) = 3x+4$ and $g(x) = f^{-1}(x) = \frac{x-4}{3}$ . Suppose we wanted to solve $3x+4 = 7$ . Going through the usual machinations, we obtain $x = 1$ .

If we view this equation as $f(x) = 7$ , however, then we are looking for the input $x$ corresponding to the output $f(x) = 7$ . This is exactly the question $f^{-1}$ was built to answer. In other words, the solution to $f(x) = 7$ is $x = f^{-1}(7) =1$ . In other words, the formula $f^{-1}(x)$ encodes all of the algebra required to `undo’ what the formula $f(x)$ does to $x$ . More generally, any time you have ever solved an equation, you have really been working through an inverse problem.

We also note the graphs of $f(x) = 3x+4$ and $g(x) = f^{-1}(x) = \frac{x-4}{3}$ are easily seen to be reflections across the line $y=x$ , as seen below. In particular, note that the $y$ -intercept $(0,4)$ on the graph of $y = f(x)$ corresponds to the $x$ -intercept on the graph of $y = f^{-1}(x)$ . Indeed, the point $(0,4)$ on the graph of $y = f(x)$ can be interpreted as $(0,4) = (0,f(0)) = (f^{-1}(4), 4)$ just as the point $(4,0)$ on the graph of $y = f^{-1}(x)$ can be interpreted as $(4,0) = (4, f^{-1}(4)) = (f(0), 0)$ .

A graph of the line y =3x+4, in red, and line y = (x-4)/3, in blue. The graph included a dashed line for y=x. The point (0,4) is on the red line and the point (4,0) is on the blue line. The graph is labeled with both functions. — Graphical representation of a function and its inverse.

Example 5.1.1

Example 5.1.1.1

For each pair of functions $f$ and $g$ below:

$f(x) = \sqrt[3]{x-1} + 2$ and $g(x) = (x-2)^3+1$

Verify each pair of functions $f$ and $g$ are inverses: (a) algebraically and (b) graphically.
Use the fact $f$ and $g$ are inverses to solve $f(x) = 5$ and $g(x) = -3$

Solution:

Solution for $f(x) = \sqrt[3]{x-1} + 2$ and $g(x) = (x-2)^3+1$ .

1. (a) To verify $f(x) = \sqrt[3]{x-1} + 2$ and $g(x) = (x-2)^3+1$ are inverses, we appeal to Definition 5.1 and show $(g \circ f)(x) = x$ and $(f \circ g)(x) = x$ for all real numbers, $x$ .
  $\begin{array}{lcl} \begin{array}{rcl} (g \circ f)(x) & = & g(f(x)) \\ & = & g(\sqrt[3]{x-1} + 2) \\ & = & [ (\sqrt[3]{x-1} + 2)-2]^3 + 1 \\ & = & (\sqrt[3]{x-1})^3 + 1 \\ & = & x-1+1 \\ & = & x \, \, \checkmark \\ \end{array} & \qquad & \begin{array}{rcl} (f \circ g)(x) & = & f(g(x)) \\ & = & f((x-2)^3+1) \\ & = & \sqrt[3]{[(x-2)^3+1] -1}+2 \\ & = & \sqrt[3]{(x-2)^3} +2\\ & = & x-4+4 \\ & = & x \, \, \checkmark \\ \end{array} \\ \end{array}$
  
  As the root here, $3$ , is odd, Theorem 4.2 gives $(\sqrt[3]{x-1})^3 = x-1$ and $\sqrt[3]{(x-2)^3} = x-2$ .
  (b) To show $f$ and $g$ are inverses graphically, we graph $y = f(x)$ and $y = g(x)$ on the same set of axes and check to see if they are reflections about the line $y=x$ .
  The graph of $y = f(x) = \sqrt[3]{x-1} + 2$ appears below on the left courtesy of Theorem 4.1 in Section 4.1. The graph of $y = g(x) = (x-2)^3+1$ appears below in the middle thanks to Theorem 2.2 in Section 2.2.
  We can immediately see three pairs of corresponding points: $(0,1)$ and $(1,0)$ , $(1,2)$ and $(2,1)$ , $(2,3)$ and $(3,2)$ . When graphed on the same pair of axes, the two graphs certainly appear to be symmetric about the line $y=x$ , as required.
2. $f$ and $g$ are inverses, so the solution to $f(x) = 5$ is $x = f^{-1}(5) = g(5) = (5-2)^3+1 = 28$ . To check, we find $f(28) = \sqrt[3]{28-1}+2 = \sqrt[3]{27} + 2 = 3+2 = 5$ , as required.
  Likewise, the solution to $g(x) = -3$ is $x = g^{-1}(-3) = f(-3) = \sqrt[3]{(-3)-1} + 2 = 2 - \sqrt[3]{4}$ . Once again, to check, we find $g(2 - \sqrt[3]{4}) = (2 - \sqrt[3]{4}-2)^3 + 1 = (-\sqrt[3]{4})^3 +1 = -4+1 = -3$ .

Example 5.1.1.2

For each pair of functions $f$ and $g$ below:

$f(t) = \dfrac{2t}{t+1}$ and $g(t) = \dfrac{t}{2-t}$

Verify each pair of functions $f$ and $g$ are inverses: (a) algebraically and (b) graphically.
Use the fact $f$ and $g$ are inverses to solve $f(x) = 5$ and $g(x) = -3$

Solution:

Solution for $f(t) = \dfrac{2t}{t+1}$ and $g(t) = \dfrac{t}{2-t}$ .

(a) Note the domain of $f$ excludes $t = -1$ and the domain of $g$ excludes $t=2$ . Hence, when simplifying $(g \circ f)(t)$ and $(f \circ g)(t)$ , we tacitly assume $t \neq -1$ and $t \neq 2$ , respectively.
$\begin{array}{ccc} \begin{array}{rcl} (g \circ f)(t) & = & g(f(t)) \\ [6pt] & = & g \left(\dfrac{2t}{t+1} \right) \\ [10pt] & = & \dfrac{\dfrac{2t}{t+1} }{2 - \dfrac{2t}{t+1}} \\ [25pt] & = & \dfrac{\dfrac{2t}{t+1} }{2 - \dfrac{2t}{t+1}} \cdot \dfrac{(t+1)}{(t+1)} \\ [25pt] & = & \dfrac{2t}{2(t+1) - 2t} \\ [10pt] & = & \dfrac{2t}{2t+2-2t} \\ [8pt] & = & \dfrac{2t}{2} \\ [8pt] & = & t \, \, \checkmark \\ \end{array} & \qquad & \begin{array}{rcl} (g \circ f)(t) & = & g(f(t)) \\ [6pt] & = & g \left(\dfrac{2t}{t+1} \right) \\ [10pt] & = & \dfrac{\dfrac{2t}{t+1} }{2 - \dfrac{2t}{t+1}} \\ [25pt] & = & \dfrac{\dfrac{2t}{t+1} }{2 - \dfrac{2t}{t+1}} \cdot \dfrac{(t+1)}{(t+1)} \\ [25pt] & = & \dfrac{2t}{2(t+1) - 2t} \\ [10pt] & = & \dfrac{2t}{2t+2-2t} \\ [8pt] & = & \dfrac{2t}{2} \\ [8pt] & = & t \, \, \checkmark \\ \end{array} \\ \end{array}$

(b) We graph $y=f(t)$ and $y=g(t)$ using the techniques discussed in Sections 3.2 and 3.3.

We find the graph of $f$ has a vertical asymptote $t=-1$ and a horizontal asymptote $y = 2$ . Corresponding to the vertical asymptote $t=-1$ on the graph of $f$ , we find the graph of $g$ has a horizontal asymptote $y=-1.$

Likewise, the horizontal asymptote $y=2$ on the graph of $f$ corresponds to the vertical asymptote $t=2$ on the graph of $g$ . Both graphs share the intercept $(0,0)$ . When graphed together on the same set of axes, the graphs of $f$ and $g$ do appear to be symmetric about the line $y=t.$
Don’t let the fact that $f$ and $g$ in this case were defined using the independent variable, ` $t$ ‘ instead of ` $x$ ‘ deter you in your efforts to solve $f(x) = 5$ . Remember that, ultimately, the function $f$ here is the process represented by the formula $f(t)$ , and is the same process (with the same inverse!) regardless of the letter used as the independent variable. Hence, the solution to $f(x) = 5$ is $x = f^{-1}(1) = g(5)$ . We get $g(5) = \frac{5}{2-5} = -\frac{5}{3}$ .
To check, we find $f\left(-\frac{5}{3} \right) = \left(-\frac{10}{3}\right) / \left(-\frac{2}{3} \right) = 5$ . Similarly, we solve $g(x) = -3$ by finding $x = g^{-1}(-3) = f(-3) = \frac{-6}{-2} = 3$ . Sure enough, we find $g(3) = \frac{3}{2-3} = -3$ .

We now investigate under what circumstances a function is invertible. As a way to motivate the discussion, we consider $f(x) = x^2$ . A likely candidate for the inverse is the function $g(x) = \sqrt{x}$ . However, $(g\circ f)(x) = g(f(x)) = \sqrt{x^2} = |x|$ , which is not equal to $x$ unless $x \geq 0$ . For example, when $x=-2$ , $f(-2)= (-2)^2 = 4$ , but $g(4) = \sqrt{4}=2$ . That is, $g$ failed to return the input $-2$ from its output $4$ . Instead, $g$ matches the output $4$ to a different input, namely $2$ , which satisfies $f(2) = 4$ . Schematically:

Two ovals, the first representing the domain of the function f and the second representing the range of the function f. There are two points x = -2 and x = 2 labeled in the first oval and 4 is a point in the second oval. There is an arrow drawn from the first oval to the second labeled f. A second arrow goes from the second oval to the first, the path is dashed, and is labeled g. There are also to solid lines, one drawn from x=-2 to 4 and another from x=2 to 4. The dashed line goes from 4 to x=2. — Graphical representation of a non-invertible function

We see from the diagram that both $f(-2)$ and $f(2)$ are $4$ , thus it is impossible to construct a function which takes $4$ back to both $x=2$ and $x=-2$ . Recall that by definition, a function can match $4$ with only one number.

In general, in order for a function to be invertible, each output can come from only one input. By definition, a function matches up each input to only one output, thus invertible functions have the property that they match one input to one output and vice-versa. We formalize this concept below.

Definition 5.2

A function $f$ is said to be one-to-one if whenever $f(a) = f(b)$ , then $a=b.$

Note that an equivalent way to state Definition 5.2 is that a function is one-to-one if different inputs go to different outputs. That is, if $a \neq b$ , then $f(a) \neq f(b).$

Before we solidify the connection between invertible functions and one-to-one functions, we take a moment to see what goes wrong graphically when trying to find the inverse of $f(x) = x^2$ .

Per Theorem 5.1, the graph of $y = f^{-1}(x)$ , if it exists, is obtained from the graph of $y=x^2$ by reflecting $y=x^2$ about the line $y=x$ . Procedurally, this is accomplished by interchanging the $x$ and $y$ coordinates of each point on the graph of $y = x^2$ . Algebraically, we are swapping the variables ` $x$ ‘ and ` $y$ ‘ which results in the equation $x = y^2$ whose graph is below on the right.

Two graphs with an arrow between them. The first graph is of y equals x squared. The second graph is of x equals y squared. The arrow between them has the words "reflect across y=x" above the arrow and "switch x and y coordinates" below the arrow — The graphs of y=x^2 and x = y^2

We see immediately the graph of $x = y^2$ fails the Vertical Line Test, Theorem 1.2. In particular, the vertical line $x=4$ intersects the graph at two points, $(4,-2)$ and $(4,2)$ meaning the relation described by $x = y^2$ matches the $x$ -value $4$ with two different $y$ -values, $-2$ and $2$ .

Note that the vertical line $x=4$ and the points $(4, \pm 2)$ on the graph of $x=y^2$ correspond to the horizontal line $y=4$ and the points $(\pm 2, 4)$ on the graph of $y = x^2$ which brings us right back to the concept of one-to-one. The fact that both $(-2,4)$ and $(2,4)$ are on the graph of $f$ means $f(-2)=f(2) = 4$ . Hence, $f$ takes different inputs, $-2$ and $2$ , to the same output, $4$ , so $f$ is not one-to-one.

Recall the Horizontal Line Test from Exercise57 in Section 1.2. Applying that result to the graph of $f$ we say the graph of $f$ `fails’ the Horizontal Line Test because the horizontal line $y=4$ intersects the graph of $y = x^2$ more than once. This means that the equation $y=x^2$ does not represent $x$ as a function of $y$ .

Said differently, the Horizontal Line Test detects when there is at least one $y$ -value ( $4$ ) which is matched to more than one $x$ -value ( $\pm 2$ ). In other words, the Horizontal Line Test can be used to detect whether or not a function is one-to-one.

So, to review, $f(x) = x^2$ is not invertible, not one-to-one, and its graph fails the Horizontal Line Test. It turns out that these three attributes: being invertible, one-to-one, and having a graph that passes the Horizontal Line Test are mathematically equivalent. That is to say if one if these things is true about a function, then they all are; it also means that, as in this case, if one of these things isn’t true about a function, then none of them are. We summarize this result in the following theorem.

Theorem 5.2 Equivalent Conditions for Invertibility

For a function $f$ , either all of the following statements are true or none of them are:

$f$ is invertible.
The graph of $f$ passes the Horizontal Line Test.^[5]

To prove Theorem 5.2, we first suppose $f$ is invertible. Then there is a function $g$ so that $g(f(x)) = x$ for all $x$ in the domain of $f$ . If $f(a) = f(b)$ , then $g(f(a)) = g(f(b))$ . As a result of $g(f(x)) = x$ , the equation $g(f(a)) = g(f(b))$ reduces to $a = b$ . We’ve shown that if $f(a) = f(b)$ , then $a = b$ , proving $f$ is one-to-one.

Next, assume $f$ is one-to-one. Suppose a horizontal line $y=c$ intersects the graph of $y = f(x)$ at the points $(a,c)$ and $(b,c)$ . This means $f(a) = c$ and $f(b) = c$ so $f(a) = f(b)$ . Because $f$ is one-to-one, means $a=b$ so the points $(a,c)$ and $(b,c)$ are actually one in the same. This establishes that each horizontal line can intersect the graph of $f$ at most once, so the graph of $f$ passes the Horizontal Line Test.

Last, but not least, suppose the graph of $f$ passes the Horizontal Line Test. Let $c$ be a real number in the range of $f$ . Then the horizontal line $y=c$ intersects the graph of $y=f(x)$ just once, say at the point $(a,c) = (a, f(a))$ . Define the mapping $g$ so that $g(c) = g(f(a)) = a$ . The mapping $g$ is a function because each horizontal line $y=c$ where $c$ is in the range of $f$ intersects the graph of $f$ only once. By construction, we have the domain of $g$ is the range of $f$ and that for all $x$ in the domain of $f$ , $g(f(x)) = x$ . We leave it to the reader to show that for all $x$ in the domain of $g$ , $f(g(x)) = x$ , too.

Hence, we’ve shown: first, if $f$ invertible, then $f$ is one-to-one; second, if $f$ is one-to-one, then the graph of $f$ passes the Horizontal Line Test; and third, if $f$ passes the Horizontal Line Test, then $f$ is invertible. Therefore if $f$ is satisfies any one of these three conditions, then $f$ must satisfy the other two.^[6]

We put this result to work in the next example.

Example 5.1.2

Example 5.1.2.1

Determine if the following functions are one-to-one:

analytically using Definition 5.2 and
graphically using the Horizontal Line Test.

For the functions that are one-to-one, graph the inverse.

$f(x) =x^2-2x+4$

Solution:

Determine if $f(x) =x^2-2x+4$ is a one-to-one function.

To determine whether or not $f$ is one-to-one analytically, we assume $f(a) = f(b)$ and work to see if we can deduce $a = b$ . As we work our way through the problem, we encounter a quadratic equation. We rewrite the equation so it equals $0$ and factor by grouping. We get $a=b$ as one possibility, but we also get the possibility that $a=2-b$ . This suggests that $f$ may not be one-to-one. Taking $b=0$ , we get $a = 0$ or $a = 2$ . We have two different inputs with the same output as $f(0) = 4$ and $f(2) = 4$ , proving $f$ is neither one-to-one nor invertible.
$\begin{array}{rcl} f(a) & = & f(b) \\ a^2 - 2a+4 & = & b^2 - 2b+4 \\ a^2 - 2a & = & b^2 - 2b \\ a^2 - b^2 - 2a + 2b & = & 0 \\ (a+b)(a-b) - 2(a-b) & = & 0 \\ (a-b)((a+b) -2) & = & 0 \\ a-b = 0 & \text{or} & a+b -2 = 0 \\ a = b & \text{or} & a = 2-b \\ \end{array}$
We note that $f$ is a quadratic function and we graph $y=f(x)$ using the techniques presented in Section 2.1. We see the graph fails the Horizontal Line Test quite often – in particular, crossing the line $y=4$ at the points $(0,4)$ and $(2,4)$ .

Example 5.1.2.2

Determine if the following functions are one-to-one:

analytically using Definition 5.2 and
graphically using the Horizontal Line Test.

For the functions that are one-to-one, graph the inverse.

$g(t) = \dfrac{2t}{1-t}$

Solution:

Determine if $g(t) = \dfrac{2t}{1-t}$ is a one-to-one function.

We begin with the assumption that $g(a) = g(b)$ for $a$ , $b$ in the domain of $g$ (That is, we assume $a \neq 1$ and $b \neq 1$ .) Through our work, we deduce $a=b$ , proving $g$ is one-to-one.
$\begin{array}{rcl} g(a) & = & g(b) \\ [3pt] \dfrac{2a}{1-a} & = & \dfrac{2b}{1-b} \\ [6pt] 2a(1-b) & = & 2b(1-a) \\ 2a - 2ab & = & 2b - 2ba \\ 2a & = & 2b \\ a & = & b \, \, \checkmark \\ \end{array}$
We graph $y=g(t)$ using the procedure outlined in Section 3.3. We find the sole intercept is $(0,0)$ with asymptotes $t=1$ and $y = -2$ . Based on our graph, the graph of $g$ appears to pass the Horizontal Line Test, verifying $g$ is one-to-one.

Because $g$ is one-to-one, $g$ is invertible. Even though we do not have a formula for $g^{-1}(t)$ , we can nevertheless sketch the graph of $y = g^{-1}(t)$ by reflecting the graph of $y=g(t)$ across $y = t$ .

Corresponding to the vertical asymptote $t=1$ on the graph of $g$ , the graph of $y = g^{-1}(t)$ will have a horizontal asymptote $y = 1$ . Similarly, the horizontal asymptote $y=-2$ on the graph of $g$ corresponds to a vertical asymptote $t = -2$ on the graph of $g^{-1}$ . The point $(0,0)$ remains unchanged when we switch the $t$ and $y$ coordinates, so it is on both the graph of $g$ and $g^{-1}$ .

Example 5.1.2.3

Determine if the following functions are one-to-one:

analytically using Definition 5.2 and
graphically using the Horizontal Line Test.

For the functions that are one-to-one, graph the inverse.

$F = \{(-1,1), (0,2), (1,-3), (2,1)\}$

Solution:

Determine if $F = \{(-1,1), (0,2), (1,-3), (2,1)\}$ is a one-to-one function.

The function $F$ is given to us as a set of ordered pairs. Recall each ordered pair is of the form $(a, F(a))$ . As $(-1,1)$ and $(2,1)$ are both elements of $F$ , this means $F(-1)=1$ and $F(2) = 1$ .
Hence, we have two distinct inputs, $-1$ and $2$ with the same output, $1$ , thus $F$ is not one-to-one and, hence, not invertible.
To graph $F$ , we plot the points in $F$ below on the left. We see the horizontal line $y=1$ crosses the graph more than once. Hence, the graph of $F$ fails the Horizontal Line Test.

Example 5.1.2.4

Determine if the following functions are one-to-one:

analytically using Definition 5.2 and
graphically using the Horizontal Line Test.

For the functions that are one-to-one, graph the inverse.

$G = \{ (t^3+1, 2t) \, | \, t \text{ is a real number.} \}$

Solution:

Determine if $G = \{ (t^3+1, 2t) \, | \, t \, \text{ is a real number} \}$ is a one-to-one function.

Like the function $F$ above, the function $G$ is described as a set of ordered pairs. Before we set about determining whether or not $G$ is one-to-one, we take a moment to show $G$ is, in fact, a function. That is, we must show that each real number input to $G$ is matched to only one output.

We are given $G = \{ (t^3+1, 2t) \, | \, t \, \text{ is a real number} \}$ . and we know that when represented in this way, each ordered pair is of the form (input, output). Hence, the inputs to $G$ are of the form $t^3+1$ and the outputs from $G$ are of the form $2t$ . To establish $G$ is a function, we must show that each input produces only one output. If it should happen that $a^3+1 = b^3+1$ , then we must show $2a = 2b$ . The equation $a^3+1 = b^3+1$ gives $a^3=b^3$ , or $a=b$ . From this it follows that $2a=2b$ so $G$ is a function.

To show $G$ is one-to-one, we must show that if two outputs from $G$ are the same, the corresponding inputs must also be the same. That is, we must show that if $2a=2b$ , then $a^3+1 = b^3+1$ . We see almost immediately that if $2a=2b$ then $a=b$ so $a^3+1 = b^3+1$ as required. This shows $G$ is one-to-one and, hence, invertible.
We graph $G$ below on the left by plotting points in the default $xy$ -plane by choosing different values for $t$ . For instance, $t=0$ corresponds to the point $(0^3+1, 2(0)) = (1,0)$ , $t=1$ corresponds to the point $(1^3+1, 2(1)) = (2,2)$ , $t=-1$ corresponds to the point $((-1)^3+1, 2(-1)) = (0, -2)$ , etc. Our graph appears to pass the Horizontal Line Test, confirming $G$ is one-to-one. We obtain the graph of $G^{-1}$ below on the right by reflecting the graph of $G$ about the line $y=x.$

In Example 5.1.2, we showed the functions $G$ and $g$ are invertible and graphed their inverses. While graphs are perfectly fine representations of functions, we have seen where they aren’t the most accurate. Ideally, we would like to represent $G^{-1}$ and $g^{-1}$ in the same manner in which $G$ and $g$ are presented to us. The key to doing this is to recall that inverse functions take outputs back to their associated inputs.

Consider $G = \{ (t^3+1, 2t) \, | \, t \,\text{ is a real number} \}$ . As mentioned in Example 5.1.2, the ordered pairs which comprise $G$ are in the form (input, output). Hence to find a compatible description for $G^{-1}$ , we simply interchange the expressions in each of the coordinates to obtain $G^{-1} = \{ (2t, t^3+1) \, | \, t \, \text{ is a real number} \}$ .

The function $g$ was defined in terms of a formula, so we would like to find a formula representation for $g^{-1}$ . We apply the same logic as above. Here, the input, represented by the independent variable $t$ , and the output, represented by the dependent variable $y$ , are related by the equation $y = g(t)$ . Hence, to exchange inputs and outputs, we interchange the ` $t$ ‘ and ` $y$ ‘ variables. Doing so, we obtain the equation $t = g(y)$ which is an implicit description for $g^{-1}$ . Solving for $y$ gives an explicit formula for $g^{-1}$ , namely $y = g^{-1}(t)$ . We demonstrate this technique below.

$\begin{array}{rclr} y & = & g(t) & \\ [5pt] y & = & \dfrac{2t}{1-t} & \\ [7pt] t & = & \dfrac{2y}{1-y} & \text{interchange variables: } t \text{ and } y \\ [3pt] t(1-y) & = & 2y & \\ [3pt] t-ty & = & 2y & \\ [3pt] t & = & ty + 2y & \\ [3pt] t & = & y(t+2) & \text{factor}\\ [8pt] y & = & \dfrac{t}{t+2} \end{array}$

We claim $g^{-1}(t) = \frac{t}{t+2}$ , and leave the algebraic verification of this to the reader.

We generalize this approach below. As always, we resort to the default ` $x$ ‘ and ` $y$ ‘ labels for the independent and dependent variables, respectively.

Steps for finding a formula for the inverse of a one-to-one function

Write $y=f(x)$
Interchange $x$ and $y$
Solve $x = f(y)$ for $y$ to obtain $y=f^{-1}(x)$

We now return to $f(x) = x^2$ . We know that $f$ is not one-to-one, and thus, is not invertible, but our goal here is to see what goes wrong algebraically.

If we attempt to follow the algorithm above to find a formula for $f^{-1}(x)$ , we start with the equation $y=x^2$ and interchange the variables ` $x$ ‘ and ` $y$ ‘ to produce the equation $x = y^2$ . Solving for $y$ gives $y = \pm \sqrt{x}.$ It’s this ` $\pm$ ‘ which is causing the problem for us as this produces two $y$ -values for any $x>0.$

Using the language of Section 1.2, the equation $x = y^2$ implicitly defines two functions, $g_{1}(x) = \sqrt{x}$ and $g_{2}(x) = -\sqrt{x}$ , each of which represents the top and bottom halves, respectively, of the graph of $x = y^2 .$

Three graphs side by side. The first graph is of a parabola opening to the right. The vertex is at (0,0) and the points (4,2) and (-4,2) are labeled on the graph. The graph is labeled x = y squared. The second graph is of the upper half of the same parabola from the first graph, including the point (4,2). The graph is labeled y equals g_1 (x) equals the square root of x. The third and final graph is of the lower half of the parabola from the first graph, including the point (4,-2). The graph is labeled y equals g_2 (x) equals the negative square root of x. — The graph of x = y^2 and its two corresponding functions

Hence, in some sense, we have two partial inverses for $f(x) = x^2$ : $g_{1}(x) = \sqrt{x}$ returns the positive inputs from $f$ and $g_{2}(x) = -\sqrt{x}$ returns the negative inputs to $f$ . In order to view each of these functions as strict inverses, however, we need to split $f$ into two parts: $f_{1}(x) = x^2$ for $x \geq 0$ and $f_{2}(x) = x^2$ for $x \leq 0$ .

Three graphs side by side. The first graph is of a parabola opening upward. The vertex is at (0,0) and the points (24) and (-2,4) are labeled on the graph. The graph is labeled y = x squared. The second graph is of the right half of the same parabola from the first graph, including the point (2,4). The graph is labeled y equals f_1 (x) equals x squared for x greater than or equal to zero. The third and final graph is of the left half of the parabola from the first graph, including the point (-2,4). The graph is labeled y equals f_2 (x) equals x squared for x less than or equal to zero. — The graphs of y =x^2 and each half of the graph

We claim that $f_{1}$ and $g_{1}$ are an inverse function pair as are $f_{2}$ and $g_{2}$ . Indeed, we find:

$\begin{array}{rcl} \begin{array}{rcl} (g_{1} \circ f_{1})(x) & = & g_{1}(f_{1}(x)) \\ & = & g_{1}(x^2) \\ & = & \sqrt{x^2} \\ & = & |x| = x, \, \text{as $x \geq 0$.} \\ \end{array} & \qquad \qquad & \begin{array}{rcl} (f_{1} \circ g_{1})(x) & = & f_{1}(g_{1}(x)) \\ & = & f_{1}(\sqrt{x}) \\ & = & (\sqrt{x})^2 \\ & = & x \\ \end{array} \\ \end{array}$

The graph includes two curves. The first curve is of y equals x squared for x greater than or equal to zero, in red. The point (2,4) is marked, but not labeled on the curve. The second curve is of y equals the square root of x, in blue. The point (4,2) is marked, but not labeled on the curve. The graph also includes the dashed line representing y equals x. The graph is labeled with the equations of both curves. — The graph of the inverse pair f_1 and g_1

$\begin{array}{rcl} \begin{array}{rcl} (g_{2} \circ f_{2})(x) & = & g_{2}(f_{2}(x)) \\ & = & g_{2}(x^2) \\ & = & -\sqrt{x^2} \\ & = & - |x| \\ & = & -(-x) = x, \, \text{as $x \leq 0$.} \end{array} & \qquad \qquad & \begin{array}{rcl} (f_{2} \circ g_{2})(x) & = & f_{2}(g_{2}(x)) \\ & = & f_{2}(-\sqrt{x}) \\ & = & (-\sqrt{x})^2 \\ & = & (\sqrt{x})^2 \\ & = & x \\ \end{array} \\ \end{array}$

The graph includes two curves. The first curve is of y equals x squared for x less than or equal to zero, in red. The point (-2,4) is marked, but not labeled on the curve. The second curve is of y equals the negative square root of x, in blue. The point (4,-2) is marked, but not labeled on the curve. The graph also includes the dashed line representing y equals x. The graph is labeled with the equations of both curves. — The graph of the inverse pair f_2 and g_2

Hence, by restricting the domain of $f$ we are able to produce invertible functions. Said differently, because the equation $x = y^2$ implicitly describes a pair of functions, the equation $y = x^2$ implicitly describes a pair of invertible functions.

Our next example continues the theme of restricting the domain of a function to find inverse functions.

Example 5.1.3

Example 5.1.3.1

Graph the following functions to show they are one-to-one and determine their inverses. Check your answers analytically using function composition and graphically.

$j(x) = x^2 - 2x + 4$ , $x \leq 1$

Solution:

Graph $j(x) = x^2 - 2x + 4$ , $x \leq 1$ to show it is one-to-one and determine its inverse.

The function $j$ is a restriction of the function $f$ from Example 5.1.2. The domain of $j$ is restricted to $x \leq 1$ , therefore we are selecting only the `left half’ of the parabola. Hence, the graph of $j$ , seen below, passes the Horizontal Line Test and thus $j$ is invertible.

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Next, we find an explicit formula for $j^{-1}(x)$ using our standard algorithm.^[7]

$\begin{array}{rclr} y & = & j(x) & \\ y & = & x^2-2x+4, \, \, \, x \leq 1 \\ x & = & y^2 - 2y+4, \, \, \, y \leq 1 & \text{switch } x \text{ and } y \\ 0 & = & y^2 - 2y + 4-x & \\ y & = & \dfrac{2 \pm \sqrt{(-2)^2-4(1)(4-x)}}{2(1)} & \text{quadratic formula, } c=4-x \\ [10pt] y & = & \dfrac{2 \pm \sqrt{4x-12}}{2} & \\ [6pt] y & = & \dfrac{2 \pm \sqrt{4(x-3)}}{2} & \\ [6pt] y & = & \dfrac{2 \pm 2\sqrt{x-3}}{2} & \\ [6pt] y & = & \dfrac{2\left(1 \pm \sqrt{x-3}\right)}{2} & \\ [6pt] y & = & 1 \pm \sqrt{x-3} & \\ y & = & 1 - \sqrt{x-3} & \text{due to the fact that } y \leq 1 \\ \end{array}$

Hence, $j^{-1}(x) = 1 - \sqrt{x-3}$ .

To check our answer algebraically, we simplify $(j^{-1} \circ j)(x)$ and $(j \circ j^{-1})(x)$ Note the importance of the domain restriction $x \leq 1$ when simplifying $(j^{-1} \circ j)(x)$ .

$\begin{array}{rcl} \left(j^{-1} \circ j \right)(x) & = & j^{-1}(j(x)) \\ & = & j^{-1}\left(x^2-2x+4\right), \, \, \, x \leq 1 \\ & = & 1 - \sqrt{\left(x^2-2x+4\right)-3} \\ & = & 1 - \sqrt{x^2-2x+1} \\ & = & 1 - \sqrt{(x-1)^2} \\ & = & 1 - |x-1| \\ & = & 1 - (-(x-1)) \, \, \text{as $x \leq 1$}\\ & = & x \, \, \checkmark \\ \end{array}$

$\begin{array}{rcl} \left(j \circ j^{-1} \right)(x) & = & j\left(j^{-1}(x)\right) \\ & = & j\left(1 - \sqrt{x-3}\right) \\ & = & \left(1 - \sqrt{x-3}\right)^2-2\left(1 - \sqrt{x-3}\right)+4 \\ & = & 1 - 2\sqrt{x-3} + \left(\sqrt{x-3}\right)^2 -2 \\ & & + \, 2\sqrt{x-3}+4 \\ & = & 1+ x-3 -2 +4 \\ & = & x \, \, \checkmark \\ \end{array}$

We graph both $j$ and $j^{-1}$ on the axes below. They appear to be symmetric about the line $y=x$ .

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Example 5.1.3.2

Graph the following functions to show they are one-to-one 2nd determine their inverses. Check your answers analytically using function composition and graphically.

$k(t) = \sqrt{t+2} - 1$

Solution:

Graph $k(t) = \sqrt{t+2} - 1$ to show it is one-to-one and determine its inverse.

Graphing $y=k(t) =\sqrt{t+2} - 1$ , we see $k$ is one-to-one,

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so we proceed to find an formula for $k^{-1}$ .

$\begin{array}{rclr} y & = & k(t) & \\ y & = & \sqrt{t+2}-1 & \\ t & = & \sqrt{y+2} - 1 & \text{switch $t$ and $y$} \\ t+1 & = & \sqrt{y+2} & \\ (t+1)^2 & = & \left(\sqrt{y+2}\right)^2 & \\ t^2 + 2t + 1 & = & y + 2 & \\ y & = & t^2 + 2t - 1 & \\ \end{array}$

We have $k^{-1}(t) = t^2+2t-1$ . Based on our experience, we know something isn’t quite right. We determined $k^{-1}$ is a quadratic function, and we have seen several times in this section that these are not one-to-one unless their domains are suitably restricted.

Theorem 5.1tells us that the domain of $k^{-1}$ is the range of $k$ . From the graph of $k$ , we see that the range is $[-1, \infty)$ , which means we restrict the domain of $k^{-1}$ to $t \geq -1$ .

We now check that this works in our compositions. Note the importance of the domain restriction, $t \geq -1$ when simplifying $(k \circ k^{-1})(t)$ .

$\begin{array}{rcl} \left(k^{-1} \circ k \right)(t) & = & k^{-1}(k(t)) \\ & = & k^{-1}\left(\sqrt{t+2}-1\right) \\ & = & \left(\sqrt{t+2}-1\right)^2 + 2\left(\sqrt{t+2}-1\right) - 1 \\ & = & \left(\sqrt{t+2}\right)^2 - 2\sqrt{t+2} + 1 \\ && + \, 2 \sqrt{t+2} - 2 - 1 \\ & = &t+2 -2 \\ & = & t \, \, \checkmark \\ \end{array}$

$\begin{array}{rcl} \left(k \circ k^{-1} \right)(t) & = & k\left( t^2+2t-1 \right), \, \, \, t \geq -1 \\ & = & \sqrt{\left(t^2+2t-1\right)+2}-1 \\ & = & \sqrt{t^2+2t+1}-1 \\ & = & \sqrt{(t+1)^2}-1 \\ & = & |t+1| -1 \\ & = & t+1 -1, \, \, \text{as $t \geq -1$} \\ & = & t \, \, \checkmark \\ \end{array}$

Graphically, everything checks out, provided that we remember the domain restriction on $k^{-1}$ means we take the right half of the parabola.

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Our last example of the section gives an application of inverse functions. Recall in Example 1.3.9 in Section 1.3.1, we modeled the demand for PortaBoy game systems as the price per system, $p(x)$ as a function of the number of systems sold, $x$ . In the following example, we find $p^{-1}(x)$ and interpret what it means.

Example 5.1.4

Example 5.1.4.1

Recall the price-demand function for PortaBoy game systems is modeled by the formula $p(x) = -1.5x + 250$ for $0 \leq x \leq 166$ where $x$ represents the number of systems sold (the demand) and $p(x)$ is the price per system, in dollars.

Explain why $p$ is one-to-one and find a formula for $p^{-1}(x)$ . State the restricted domain.

Solution:

Explain why $p$ is one-to-one and find a formula for $p^{-1}(x)$ . State the restricted domain.

Recall the graph of $p(x) = -1.5x + 250$ , $0 \leq x \leq 166$ , is a line segment from $(0,250)$ to $(166,1)$ , and as such passes the Horizontal Line Test. Hence, $p$ is one-to-one.

We determine the expression for $p^{-1}(x)$ as usual and get $p^{-1}(x) = \frac{500-2x}{3}.$

The domain of $p^{-1}$ should match the range of $p$ , which is $[1,250]$ , and as such, we restrict the domain of $p^{-1}$ to $1 \leq x \leq 250.$

Example 5.1.4.2

Recall the price-demand function for PortaBoy game systems is modeled by the formula $p(x) = -1.5x + 250$ for $0 \leq x \leq 166$ where $x$ represents the number of systems sold (the demand) and $p(x)$ is the price per system, in dollars.

Compute and interpret $p^{-1}(220)$ .

Solution:

Compute and interpret $p^{-1}(220)$ .

We find $p^{-1}(220) = \frac{500-2(220)}{3} = 20$ .

The function $p$ took as inputs the number of systems sold and returned the price per system as the output, thus $p^{-1}$ takes the price per system as its input and returns the number of systems sold as its output.

Hence, $p^{-1}(220) = 20$ means 20 systems will be sold if the price is set at 220 dollars per system.

Example 5.1.4.3

Recall the price-demand function for PortaBoy game systems is modeled by the formula $p(x) = -1.5x + 250$ for $0 \leq x \leq 166$ where $x$ represents the number of systems sold (the demand) and $p(x)$ is the price per system, in dollars.

Recall from Section 2.1that the profit $P$ , in dollars, as a result of selling $x$ systems is given by $P(x)= -1.5x^2+170x-150$ . Write and interpret $\left( P \circ p^{-1}\right)(x)$ .

Solution:

Write and interpret $\left( P \circ p^{-1}\right)(x)$ .

We compute

$\begin{array}{rcl} \left( P \circ p^{-1}\right)(x) & = & P \left(p^{-1}(x)\right) \\ & = & P\left(\frac{500-2x}{3}\right) \\ & = & -1.5\left(\frac{500-2x}{3}\right)^2+170\left(\frac{500-2x}{3}\right)-150 \end{array}$

After a hefty amount of algebra,^[8] we obtain $\left( P \circ p^{-1}\right)(x) = -\frac{2}{3} x^2 +220x - \frac{40450}{3}$ .

To understand what this means, recall that the original profit function $P$ gave us the profit as a function of the number of systems sold. The function $p^{-1}$ gives us the number of systems sold as a function of the price.

Hence, when we compute $(P \circ p^{-1})(x) = P(p^{-1}(x))$ , we input a price per system, $x$ into the function $p^{-1}$ .

The number $p^{-1}(x)$ is the number of systems sold at that price. This number is then fed into $P$ to return the profit obtained by selling $p^{-1}(x)$ systems.

Hence, $\left(P \circ p^{-1}\right)(x)$ gives us the profit (in dollars) as a function of the price per system, $x$ .

Example 5.1.4.4

Recall the price-demand function for PortaBoy game systems is modeled by the formula $p(x) = -1.5x + 250$ for $0 \leq x \leq 166$ where $x$ represents the number of systems sold (the demand) and $p(x)$ is the price per system, in dollars.

Use your answer to part 3 to determine the price per PortaBoy which would yield the maximum profit. Compare with Example 2.1.3.

Solution:

Use your answer to part 3 to determine the price per PortaBoy which would yield the maximum profit. Compare with Example 2.1.3.

We know from Section 2.1 that the graph of $y = \left( P \circ p^{-1}\right)(x)$ is a parabola opening downwards. The maximum profit is realized at the vertex. We are concerned only with the price per system, so we need only find the $x$ -coordinate of the vertex.

Identifying $a = -\frac{2}{3}$ and $b = 220$ , we get, by the Vertex Formula, Equation 2.1, $x = -\frac{b}{2a} = 165$ .

Hence, the weekly profit is maximized if we set the price at 165 dollars per system. Comparing this with our answer from Example 2.1.3, there is a slight discrepancy to the tune of 50 cents. We leave it to the reader to balance the books appropriately.

5.1.1 Section Exercises

In Exercises 1 – 8, verify the given pairs of functions are inverses algebraically and graphically.

$f(x) = 2x+7$ and $g(x) = \dfrac{x-7}{2}$
$f(x) = \dfrac{5-3x}{4}$ and $g(x) = -\dfrac{4}{3} x + \dfrac{5}{3}$
$f(t) = \dfrac{5}{t-1}$ and $g(t) = \dfrac{t+5}{t}$
$f(t) = \dfrac{t}{t-1}$ and $g(t) = f(t) = \dfrac{t}{t-1}$
$f(x) = \sqrt{4-x}$ and $g(x) = -x^2+4$ , $x \geq 0$
$f(x) = 1-\sqrt{x+1}$ and $g(x) = x^2-2x$ , $x \leq 1$
$f(t) = (t-1)^3+5$ and $g(t) = \sqrt[3]{t-5}+1$
$f(t) = -\sqrt[4]{t-2}$ and $g(t) = t^4+2$ , $t \leq 0$

In Exercises 9 – 28, show that the given function is one-to-one and find its inverse. Check your answers algebraically and graphically. Verify the range of the function is the domain of its inverse and vice-versa.

$f(x) = 6x - 2$
$f(x) = 42-x$
$g(t) = \dfrac{t-2}{3} + 4$
$g(t) = 1 - \dfrac{4+3t}{5}$
$f(x) = \sqrt{3x-1}+5$
$f(x) = 2-\sqrt{x - 5}$
$g(t) = 3\sqrt{t-1}-4$
$g(t) = 1 - 2\sqrt{2t+5}$
$f(x) = \sqrt[5]{3x-1}$
$f(x) = 3-\sqrt[3]{x-2}$
$g(t) = t^2 - 10t$ , $t \geq 5$
$g(t) = 3(t + 4)^{2} - 5, \; t \leq -4$
$f(x) = x^2-6x+5, \; x \leq 3$
$f(x) = 4x^2 + 4x + 1$ , $x < -1$
$g(t) = \dfrac{3}{4-t}$
$g(t) = \dfrac{t}{1-3t}$
$f(x) = \dfrac{2x-1}{3x+4}$
$f(x) = \dfrac{4x + 2}{3x - 6}$
$g(t) = \dfrac{-3t - 2}{t + 3}$
$g(t) = \dfrac{t-2}{2t-1}$
Explain why each set of ordered pairs below represents a one-to-one function and find the inverse.
1. $F = \{ (0,0), (1,1), (2,-1), (3,2), (4,-2), (5,3), (6,-3) \}$
2. $G = \{ (0,0), (1,1), (2,-1), (3,2), (4,-2), (5,3), (6,-3), \ldots \}$ NOTE: The difference between $F$ and $G$ is the ` $\ldots$ .’
3. $P = \{ (2t^5, 3t-1) \, | \, t \, \text{ is a real number} \}$
4. $Q = \{ (n, n^2) \, | \, n \, \text{ is a \textit{natural} number} \}$ ^[9]

In Exercises 30 – 33, explain why each graph represents^[10] a one-to-one function and graph its inverse.

$y=f(x)$
$y=g(t)$
$y=S(t)$
$y=R(s)$
The price of a dOpi media player, in dollars per dOpi, is given as a function of the weekly sales according to the formula for .
1. Find $p^{-1}(x)$ and state its domain.
2. Compute and interpret $p^{-1}(105)$ .
3. The profit (in dollars) made from producing and selling $x$ dOpis per week is given by the formula $P(x)= -15x^2+350x-2000$ , for $0 \leq x \leq 30$ . Find $\left(P \circ p^{-1}\right)(x)$ and determine what price per dOpi would yield the maximum profit. What is the maximum profit? How many dOpis need to be produced and sold to achieve the maximum profit?
Show that the Fahrenheit to Celsius conversion function found in Exercise 67 in Section1.3.3 is invertible and that its inverse is the Celsius to Fahrenheit conversion function.
Analytically show that the function $f(x) = x^3 + 3x + 1$ is one-to-one. Use Theorem 5.1 to help you compute $f^{-1}(1), \; f^{-1}(5), \;$ and $f^{-1}(-3)$ . What happens when you attempt to find a formula for $f^{-1}(x)$ ?
Let .
1. Graph $y = f(x)$ using the techniques in Section 3.3. Check your answer using a graphing utility.
2. Verify that $f$ is one-to-one on the interval $(-1,1)$ .
3. Use the procedure outlined in this section to find the formula for $f^{-1}(x)$ for $-1 < x < 1$ .
4. Because $f(0) = 0$ , it should be the case that $f^{-1}(0) = 0$ . What goes wrong when you attempt to substitute $x=0$ into $f^{-1}(x)$ ? Discuss with your classmates how this problem arose and possible remedies.
The function given in number 4 is an example of a function which is its own inverse.
1. Algebraically verify every function of the form: $f(x) = \dfrac{ax + b}{cx - a}$ is its own inverse.What assumptions do you need to make about the values of $a$ , $b$ , and $c$ ?
2. Under what conditions is $f(x) = mx + b$ , $m \neq 0$ its own inverse? Prove your answer.

Section 5.1 Exercise Answers can be found in the Appendix … Coming soon

At the level of functions, $g \circ f = f \circ g = I$ , where $I$ is the identity function as defined as $I(x) = x$ for all real numbers, $x$ . ↵
The identity function $I$ , first introduced in Exercise 76 in Section 1.3.1 and mentioned in Theorem 1.6, has a domain of all real numbers. As the domains of $f$ and $g$ may not be all real numbers, we need the restrictions listed here. ↵
In other words, invertible functions have exactly one inverse. ↵
See Example 1.1.5 in Section 1.1 and Example 1.35 in Section 1.3.1. ↵
i.e., no horizontal line intersects the graph more than once. ↵
For example, if we know $f$ is one-to-one, we showed the graph of $f$ passes the HLT which, in turn, guarantees $f$ is invertible. ↵
Here, we use the Quadratic Formula to solve for $y$ . For `completeness,' we note you can (and should!) also consider solving for $y$ by `completing' the square. ↵
It is good review to actually do this! ↵
Recall this means $n = 0, 1, 2, \ldots$ ↵
or, more precisely, appears to represent $\ldots$ ↵

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Functions, Trigonometry, and Systems of Equations Copyright © by Vanessa Coffelt is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted.

Definition 5.1

Example 5.1.1

Solution:

Solution:

Definition 5.2

Example 5.1.2

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Solution:

Steps for finding a formula for the inverse of a one-to-one function

Example 5.1.3

Solution:

Solution:

Example 5.1.4

Solution:

Solution:

Solution:

Solution:

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