1.7: Inverse Functions

Motivating Questions

• What does it mean to say that a given function has an inverse function?
• How can we determine whether or not a given function has a corresponding inverse function?
• When a function has an inverse function, what important properties does the inverse function have in comparison to the original function?

Because every function is a process that converts a collection of inputs to a corresponding collection of outputs, a natural question is: for a particular function, can we change perspective and think of the original function's outputs as the inputs for a reverse process? If we phrase this question algebraically, it is analogous to asking: given an equation that defines \(y\) is a function of \(x\text{,}\) is it possible to find a corresponding equation where \(x\) is a function of \(y\text{?}\)

Preview Activity \(\PageIndex{1}\)

Recall that \(F = g(C) = \frac{9}{5}C + 32\) is the function that takes Celsius temperature inputs and produces the corresponding Fahrenheit temperature outputs.

1. Show that it is possible to solve the equation \(F = \frac{9}{5}C + 32\) for \(C\) in terms of \(F\) and that doing so results in the equation \(C = \frac{5}{9}(F-32)\text{.}\)
2. Note that the equation \(C = \frac{5}{9}(F-32)\) expresses \(C\) as a function of \(F\text{.}\) Call this function \(h\) so that \(C = h(F) = \frac{5}{9}(F-32)\text{.}\)

   Find the simplest expression that you can for the composite function \(j(C) = h(g(C))\text{.}\)

3. Find the simplest expression that you can for the composite function \(k(F) = g(h(F))\text{.}\)
4. Why are the functions \(j\) and \(k\) so simple? Explain by discussing how the functions \(g\) and \(h\) process inputs to generate outputs and what happens when we first execute one followed by the other.

When a function has an inverse function

In Preview Activity 1.7.1, we found that for the function \(F = g(C) = \frac{9}{5}C + 32\text{,}\) it's also possible to solve for \(C\) in terms of \(F\) and write \(C = h(F) = \frac{5}{9}(C-32)\text{.}\) The first function, \(g\text{,}\) converts Celsius temperatures to Fahrenheit ones; the second function, \(h\text{,}\) converts Fahrenheit temperatures to Celsius ones. Thus, the process \(h\) reverses the process of \(g\text{,}\) and likewise the process of \(g\) reverses the process of \(h\text{.}\) This is also why it makes sense that \(h(g(C)) = C\) and \(g(h(F)) = F\text{.}\) If, for instance, we take a Celsius temperature \(C\text{,}\) convert it to Fahrenheit, and convert the result back to Celsius, we arrive back at the Celsius temperature we started with: \(h(g(C)) = C\text{.}\)

Similar work is sometimes possible with other functions. When we can find a new function that reverses the process of the original function, we say that the original function “has an inverse function” and make the following formal definition.

Definition \(\PageIndex{1}\)

Let \(f : A \to B\) be a function. If there exists a function \(g : B \to A\) such that

\[ g(f(a)) = a \text{ and } f(g(b)) = b \nonumber \]

for each \(a\) in \(A\) and each \(b\) in \(B\text{,}\) then we say that \(f\) has an inverse function and that the function \(g\) is the inverse of \(f\).

Note particularly what the equation \(g(f(a)) = a\) says: for any input \(a\) in the domain of \(f\text{,}\) the function \(g\) will reverse the process of \(f\) (which converts \(a\) to \(f(a)\)) because \(g\) converts \(f(a)\) back to \(a\text{.}\)

When a given function \(f\) has a corresponding inverse function \(g\text{,}\) we usually rename \(g\) as \(f^{-1}\text{,}\) which we read aloud as “\(f\)-inverse”. The equation \(g(f(a))=a\) now reads \(f^{-1}(f(a)) = a\text{,}\) which we interpret as saying “\(f\)-inverse converts \(f(a)\) back to \(a\)”. We similarly write that \(f(f^{-1}(b)) = b\text{.}\)

Activity \(\PageIndex{2}\)

Recall Dolbear's function \(F = D(N) = 40 + \frac{1}{4}N\) that converts the number, \(N\text{,}\) of snowy tree cricket chirps per minute to a corresponding Fahrenheit temperature. We have earlier established that the domain of \(D\) is \([40,180]\) and the range of \(D\) is \([50,85]\text{,}\) as seen in Figure \(\PageIndex{3}\)

1. Solve the equation \(F = 40 + \frac{1}{4}N\) for \(N\) in terms of \(F\text{.}\) Call the resulting function \(N = E(F)\text{.}\)

2. Explain in words the process or effect of the function \(N = E(F)\text{.}\) What does it take as input? What does it generate as output?

3. Use the function \(E\) that you found in (a.) to compute \(j(N) = E(D(N))\text{.}\) Simplify your result as much as possible. Do likewise for \(k(F) = D(E(F))\text{.}\) What do you notice about these two composite functions \(j\) and \(k\text{?}\)

4. Consider the equations \(F = 40 + \frac{1}{4}N\) and \(N = 4(F-40)\text{.}\) Do these equations express different relationships between \(F\) and \(N\text{,}\) or do they express the same relationship in two different ways? Explain.

When a given function has an inverse function, it allows us to express the same relationship from two different points of view. For instance, if \(y = f(t) = 2t+1\text{,}\) we can show¹ that the function \(t = g(y) = \frac{y-1}{2}\) reverses the effect of \(f\) (and vice versa), and thus \(g = f^{-1}\text{.}\) We observe that

\[ y = f(t) = 2t + 1 \text{ and } t = f^{-1}(y) = \frac{y-1}{2} \nonumber \]

are equivalent forms of the same equation, and thus they say the same thing from two different perspectives. The first version of the equation is solved for \(y\) in terms of \(t\text{,}\) while the second equation is solved for \(t\) in terms of \(y\text{.}\) This important principle holds in general whenever a function has an inverse function.

Two perspectives from a function and its inverse function.

If \(y = f(t)\) has an inverse function, then the equations

\[ y = f(t) \text{ and } t = f^{-1}(y) \nonumber \]

say the exact same thing but from two different perspectives.

Determining whether a function has an inverse function

It's important to note in Definition \(\PageIndex{1}\) that we say “If there exists \(\ldots\text{.}\)” That is, we don't guarantee that an inverse function exists for a given function. Thus, we might ask: how can we determine whether or not a given function has a corresponding inverse function? As with many questions about functions, there are often three different possible ways to explore such a question: through a table, through a graph, or through an algebraic formula.

Example \(\PageIndex{2}\)

Do the functions \(f\) and \(g\) defined by Table \(\PageIndex{3}\) and Table \(\PageIndex{4}\) have corresponding inverse functions? Why or why not?

Table \(\PageIndex{3}\) The table that defines the function \(f\text{.}\)

\(x\)	\(0\)	\(1\)	\(2\)	\(3\)	\(4\)
\(f(x)\)	\(6\)	\(4\)	\(3\)	\(4\)	\(6\)

Table \(\PageIndex{4}\) The table that defines the function \(g\text{.}\)

\(x\)	\(0\)	\(1\)	\(2\)	\(3\)	\(4\)
\(g(x)\)	\(3\)	\(1\)	\(4\)	\(2\)	\(0\)

Solution

For any function, the question of whether or not it has an inverse comes down to whether or not the process of the function can be reliably reversed. For functions given in table form such as \(f\) and \(g\text{,}\) we essentially ask if it's possible to swich the input and output rows and have the new resulting table also represent a function.

The function \(f\) does not have an inverse function because there are two different inputs that lead to the same output: \(f(0) = 6\) and \(f(4) = 6\text{.}\) If we attempt to reverse this process, we have a situation where the input \(6\) would correspond to two potential outputs, \(4\) and \(6\text{.}\)

However, the function \(g\) does have an inverse function because when we reverse the rows in Table \(\PageIndex{4}\), each input (in order, \(3\text{,}\) \(1\text{,}\) \(4\text{,}\) \(2\text{,}\) \(0\)) indeed corresponds to one and only one output (in order, \(0\text{,}\) \(1\text{,}\) \(2\text{,}\) \(3\text{,}\) \(4\)). We can thus make observations such as \(g^{-1}(4) = 2\text{,}\) which is the same as saying that \(g(2) = 4\text{,}\) just from a different perspective.

In Example \(\PageIndex{2}\), we see that if we can identify one pair of distinct inputs that lead to the same output (such as \(f(0) = f(4) = 6\) in Table 1.7.3), then the process of the function cannot be reversed and the function does not have an inverse.

Example \(\PageIndex{5}\)

Do the functions \(p\) and \(q\) defined by Figure \(\PageIndex{6}\) and Figure \(\PageIndex{7}\) have corresponding inverse functions? Why or why not?

Figure \(\PageIndex{6}\) The graph that defines function \(p\text{.}\)

Figure \(\PageIndex{7}\) The graph that defines function \(p\text{.}\)

Solution

Recall that when a point such as \((a,c)\) lies on the graph of a function \(p\text{,}\) this means that the input \(x = a\text{,}\) which represents to a value on the horizontal axis, corresponds with the output \(y = c\) that is represented by a value on the vertical axis. In this situation, we write \(p(a) = c\text{.}\) We note explicitly that \(p\) is a function because its graph passes the Vertical Line Test: any vertical line intersects the graph of \(p\) exactly, and thus each input from the domain corresponds to one and only one output.

If we attempt to change perspective and use the graph of \(p\) to view \(x\) as a function of \(y\text{,}\) we see that this fails because the output value \(c\) is associated with two different inputs, \(a\) and \(b\text{.}\) Said differently, because the horizontal line \(y = c\) intersects the graph of \(p\) at both \((a,c)\) and \((b,c)\) (as shown in Figure \(\PageIndex{6}\)), we cannot view \(y\) as the input to a function process that produces the corresonding \(x\)-value. Therefore, \(p\) does not have an inverse function.

On the other hand, provided that the behavior seen in the figure continues, the function \(q\) does have an inverse because we can view \(x\) as a function of \(y\) via the graph given in Figure \(\PageIndex{7}\) This is because for any choice of \(y\text{,}\) there corresponds one and only one \(x\) that results from \(y\text{.}\) We can think of this visually by starting at a value such as \(y = c\) on the \(y\)-axis, moving horizontally to where the line intersects the graph of \(p\text{,}\) and then moving down to the corresonding location (here \(x = a\)) on the horizontal axis. From the behavior of the graph of \(q\) (a straight line that is always increasing), we see that this correspondence will hold for any choice of \(y\text{,}\) and thus indeed \(x\) is a function of \(y\text{.}\) From this, we can say that \(q\) indeed has an inverse function. We thus can write that \(q^{-1}(c) = a\text{,}\) which is a different way to express the equivalent fact that \(q(a) = c\text{.}\)

The graphical observations that we made for the function \(q\) in Example 1.7.5 provide a general test for whether or not a function given by a graph has a corresponding inverse function.

Horizontal Line Test.

A function whose graph lies in the \(x\)-\(y\) plane has a corresponding inverse function if and only if every horizontal line intersects the graph at most once. When the graph passes this test, the horizontal coordinate of each point on the graph can be viewed as a function of the vertical coordinate of the point.

Example \(\PageIndex{8}\)

Do the functions \(r\) and \(s\) defined by

\[ y = r(t) = 3 - \frac{1}{5}(t-1)^3 \text{ and } y = s(t) = 3 - \frac{1}{5}(t-1)^2 \nonumber \]

have corresponding inverse functions? If not, use algebraic reasoning to explain why; if so, demonstrate by using algebra to find a formula for the inverse function.

Solution

For any function of the form \(y = f(t)\text{,}\) one way to determine if we can view the original input variable \(t\) as a function of the original output variable \(y\) is to attempt to solve the equation \(y = f(t)\) for \(t\) in terms of \(y\text{.}\)

Taking \(y = 3 - \frac{1}{5}(t-1)^3\text{,}\) we try to solve for \(t\) by first subtracting \(3\) from both sides to get

\[ y - 3 = -\frac{1}{5}(t-1)^3\text{.} \nonumber \]

Next, multiplying both sides by \(-5\text{,}\) it follows that

\[ (t-1)^3 = -5(y-3)\text{.} \nonumber \]

Because the cube root function has the property that \(\sqrt[3]{z^3} = z\) for every real number \(z\) (since the cube root function is the inverse function for the cubing function, and each function has both a domain and range of all real numbers), we can take the cube root of both sides of the preceding equation to get

\[ t - 1 = \sqrt[3]{-5(y-3)}\text{.} \nonumber \]

Finally, adding \(1\) to both sides, we have determined that

\[ t = 1 + \sqrt[3]{-5(y-3)}\text{.} \nonumber \]

Because we have been able to express \(t\) as a single function of \(y\) for every possible value of \(y\text{,}\) this shows that \(r\) indeed has an inverse and that \(t = r^{-1}(y) = 1 + \sqrt[3]{-5(y-3)}\text{.}\)

We attempt similar reasoning for the second function, \(y = 3 - \frac{1}{5}(t-1)^2\text{.}\) To solve for \(t\text{,}\) we first subtract \(3\) from both sides, so that

\[ y - 3 = -\frac{1}{5}(t-1)^2\text{.} \nonumber \]

After multiplying both sides by \(-5\text{,}\) we have

\[ (t-1)^2 = -5(y-3)\text{.} \nonumber \]

Next, it's necessary to take the square root of both sides in an effort to isolate \(t\text{.}\) Here, however, we encounter a crucial issue. Because the function \(g(x) = x^2\) takes any nonzero number and its opposite to the same output (e.g. \((-5)^2 = 25 = (5)^2\)), this means that we have to account for both possible inputs that result in the same output. Based on our last equation, this means that either

\[ t-1 = \sqrt{-5(y-3)} \ \text{ or } \ t-1 = -\sqrt{-5(y-3)}\text{.} \nonumber \]

As such, we find not a single equation that expresses \(t\) as a function of \(y\text{,}\) but rather two:

\[ t = 1 + \sqrt{-5(y-3)} \ \text{ or } \ t = 1 -\sqrt{-5(y-3)}\text{.} \nonumber \]

Since it appears that \(t\) can't be expressed as a single function of \(y\text{,}\) it seems to follow that \(y = s(t) = 3 - \frac{1}{5}(t-1)^2\) does not have an inverse function.

The graphs of \(y = r(t) = 3 - \frac{1}{5}(t-1)^3\) and \(y = s(t) = 3 - \frac{1}{5}(t-1)^2\) provide a different perspective to confirm the results of Example \(\PageIndex{8}\) Indeed, in Figure \(\PageIndex{9}\), we see that \(r\) appears to pass the horizontal line test because it is decreasing², and thus has an inverse function. On the other hand, the graph of \(s\) fails the horizontal line test (picture the line \(y = 2\) in Figure \(\PageIndex{10}\)) and therefore \(s\) does not have an inverse function.

Figure \(\PageIndex{9}\) A plot of \(y = r(t) = 3 - \frac{1}{5}(t-1)^3\text{.}\)

Figure \(\PageIndex{10}\) A plot of \(y = s(t) = 3 - \frac{1}{5}(t-1)^2\text{.}\)

Activity \(\PageIndex{3}\)

Determine, with justification, whether each of the following functions has an inverse function. For each function that has an inverse function, give two examples of values of the inverse function by writing statements such as “\(s^{-1}(3) = 1\)”.

1. The function \(f : S \to S\) given by Table \(\PageIndex{11}\), where \(S = \{0, 1, 2, 3, 4 \}\text{.}\)

Table \(\PageIndex{11}\). Values of \(y = f(x)\text{.}\)

\(x\)	0	1	2	3	4
\(f(x)\)	1	2	4	3	2

2. The function \(g : S \to S\) given by Table \(\PageIndex{12}\), where \(S = \{0, 1, 2, 3, 4 \}\text{.}\)

Table \(\PageIndex{12}\) Values of \(y = g(x)\text{.}\)

\(x\)	0	1	2	3	4
\(f(x)\)	4	0	3	1	2

3. The function \(p\) given by \(p(t) = 7 - \frac{3}{5}t\text{.}\) Assume that the domain and codomain of \(p\) are both “all real numbers”.

4. The function \(q\) given by \(q(t) = 7 - \frac{3}{5}t^4\text{.}\) Assume that the domain and codomain of \(q\) are both “all real numbers”.

5. The functions \(r\) and \(s\) given by the graphs in Figure \(\PageIndex{13}\) and Figure \(\PageIndex{14}\). Assume that the graphs show all of the important behavior of the functions and that the apparent trends continue beyond what is pictured.

Figure \(\PageIndex{13}\) The graph of \(y = r(t)\text{.}\)

Figure \(\PageIndex{14}\) The graph of \(y = s(t)\text{.}\)

Properties of an inverse function

When a function has an inverse function, we have observed several important relationships that hold between the original function and the corresponding inverse function.

Properties of an inverse function.

Let \(f : A \to B\) be a function whose domain is \(A\) and whose range is \(B\) be such that \(f\) has an inverse function, \(f^{-1}\text{.}\) Then:

• \(f^{-1} : B \to A\text{,}\) so the domain of \(f^{-1}\) is \(B\) and its range is \(A\text{.}\)
• The functions \(f\) and \(f^{-1}\) reverse one anothers' processes. Symbolically, \(f^{-1}(f(a)) = a\) for every input \(a\) in the domain of \(f\text{,}\) and similarly, \(f(f^{-1}(b)) = b\) for every input \(b\) in the domain of \(f^{-1}\text{.}\)
• If \(y = f(t)\text{,}\) then we can express the exact same relationship from a different perspective by writing \(t = f^{-1}(y)\text{.}\)
• Consider the setting where \(A\) and \(B\) are collections of real numbers. If a point \((x,y)\) lies on the graph of \(f\text{,}\) then it follows \(y = f(x)\text{.}\) From this, we can equivalently say that \(x = f^{-1}(y)\text{.}\) Hence, the point \((y,x)\) lies on the graph of \(x = f^{-1}(y)\text{.}\)

The last item above leads to a special relationship between the graphs of \(f\) and \(f^{-1}\) when viewed on the same coordinate axes. In that setting, we need to view \(x\) as the input of each function (since it's the horizontal coordinate) and \(y\) as the output. If we know a particular input-output relationship for \(f\text{,}\) say \(f(-1) = \frac{1}{2}\text{,}\) then it follows that \(f^{-1} \left( \frac{1}{2} \right) = -1\text{.}\) We observe that the points \(\left(-1, \frac{1}{2} \right)\) and \(\left(\frac{1}{2}, -1 \right)\) are reflections of each other across the line \(y = x\text{.}\) Because such a relationship holds for every point \((x,y)\) on the graph of \(f\text{,}\) this means that the graphs of \(f\) and \(f^{-1}\) are reflections of one another across the line \(y = x\text{,}\) as seen in Figure \(\PageIndex{15}\)

Figure \(\PageIndex{15}\) The graph of a function \(f\) along with its inverse, \(f^{-1}\text{.}\)

Activity \(\PageIndex{4}\)

During a major rainstorm, the rainfall at Gerald R. Ford Airport is measured on a frequent basis for a \(10\)-hour period of time. The following function \(g\) models the rate, \(R\text{,}\) at which the rain falls (in cm/hr) on the time interval \(t = 0\) to \(t = 10\text{:}\)

\[ R = g(t) = \frac{4}{t+2} + 1 \nonumber \]

1. Compute \(g(3)\) and write a complete sentence to explain its meaning in the given context, including units.
2. Compute the average rate of change of \(g\) on the time interval \([3,5]\) and write two careful complete sentences to explain the meaning of this value in the context of the problem, including units. Explicitly address what the value you compute tells you about how rain is falling over a certain time interval, and what you should expect as time goes on.
3. Plot the function \(y = g(t)\) using a computational device. On the domain \([1,10]\text{,}\) what is the corresponding range of \(g\text{?}\) Why does the function \(g\) have an inverse function?
4. Determine \(g^{-1} \left( \frac{9}{5} \right)\) and write a complete sentence to explain its meaning in the given context.
5. According to the model \(g\text{,}\) is there ever a time during the storm that the rain falls at a rate of exactly \(1\) centimeter per hour? Why or why not? Provide an algebraic justification for your answer.

Summary

• A given function \(f : A \to B\) has an inverse function whenever there exists a related function \(g : B \to A\) that reverses the process of \(f\text{.}\) Formally, this means that \(g\) must satisfy \(g(f(a)) = a\) for every \(a\) in the domain of \(f\text{,}\) and \(f(g(b)) = b\) for every \(b\) in the range of \(f\text{.}\)
• We determine whether or not a given function \(f\) has a corresponding inverse function by determining if the process that defines \(f\) can be reversed so that we can also think of the outputs as a function of the inputs. If we have a graph of the function \(f\text{,}\) we know \(f\) has an inverse function if the graph passes the Horizontal Line Test. If we have a formula for the function \(f\text{,}\) say \(y = f(t)\text{,}\) we know \(f\) has an inverse function if we can solve for \(t\) and write \(t = f^{-1}(y)\text{.}\)
• A good summary of the properties of an inverse function is provided in the Properties of an inverse function.

Exercises

1.

2.

3.

4.

5.

6.

7.

Consider the functions \(p\) and \(q\) whose graphs are given by Figure \(\PageIndex{16}\)

Figure \(\PageIndex{16}\) Plots of the graphs of \(p\) and \(q\text{.}\)

1. Compute each of the following values exactly, or explain why they are not defined: \(p^{-1}(2.5)\text{,}\) \(p^{-1}(-2)\text{,}\) \(p^{-1}(0)\text{,}\) and \(q^{-1}(2)\text{.}\)
2. From your work in (a), you know that the point \((2.5, -3.5)\) lies on the graph of \(p^{-1}\text{.}\) In addition to the other two points you know from \((a)\text{,}\) find three additional points that lie on the graph of \(p^{-1}\text{.}\)
3. On Figure 1.7.16, plot the \(6\) points you have determined in (a) and (b) that lie on the graph of \(y = p^{-1}(x)\text{.}\) Then, sketch the complete graph of \(y = p^{-1}(x)\text{.}\) How are the graphs of \(p\) and \(p^{-1}\) related to each other?

8.

Consider an inverted conical tank that is being filled with water. The tank's radius is \(2\) m and its depth is \(4\) m. Suppose the tank is initially empty and is being filled in such a way that the height of the water is always rising at a rate of \(0.25\) meters per minute.

Figure \(\PageIndex{17}\) The conical tank.

Figure \(\PageIndex{18}\) Axes to plot \(V = g(t)\text{.}\)

1. Explain why the height, \(h\text{,}\) of the water can be viewed as a function of \(t\) according to the formula \(h= f(t) = 0.25t\text{.}\)
2. At what time is the water in the tank \(2.5\) m deep? At what time is the tank completely full?
3. Suppose we think of the volume, \(V\text{,}\) of water in the tank as a function of \(t\) and name the function \(V = g(t)\text{.}\) Do you expect that the function \(g\) has an inverse function? Why or why not?
4. Recall that the volume of a cone of radius \(r\) and height \(h\) is \(V = \frac{\pi}{3} r^2 h\text{.}\) Due to the shape of the tank, similar triangles tell us that \(r\) and \(h\) satisfy the proportion \(r = \frac{1}{2}h\text{,}\) and thus
   \[ V = \frac{\pi}{3} \left( \frac{1}{2}h \right)^2 h = \frac{\pi}{12}h^3 \text{.}\label{E-ez-changing-inverse-V-h}\tag{1.7.1} \]

   Use the fact that \(h = f(t) = 0.25t\) along with Equation (1.7.1) to find a formula for \(V = g(t)\text{.}\) Sketch a plot of \(V = g(t)\) on the blank axes provided in Figure 1.7.18. Write at least one sentence to explain why \(V = g(t)\) has the shape that it does.

5. Take the formula for \(V = g(t)\) that you determined in (d) and solve for \(t\) to determine a formula for \(t = g^{-1}(V)\text{.}\) What is the meaning of the formula you find?
6. Find the exact time that there is \(\frac{8}{3}\pi\) cubic meters of volume in the tank.

9.

Recall that in Activity 1.6.3, we showed that Celsius temperature is a function of the number of chirps per minute from a snowy tree cricket according to the formula

\[ C = H(N) = \frac{40}{9} + \frac{5}{36}N\text{.}\label{E-ez-changing-inverse-Dolbear-C}\tag{1.7.2} \]

1. What familiar type of function is \(H\text{?}\) Why must \(H\) have an inverse function?
2. Determine an algebraic formula for \(N = H^{-1}(C)\text{.}\) Clearly show your work and thinking.
3. What is the meaning of the statement \(72 = H^{-1}\left(\frac{130}{9}\right)\text{?}\)
4. Determine the average rate of change of \(H\) on the interval \([40,50]\text{.}\) Write a complete sentence to explain the meaning of the value you find, including units on the value. Explain clearly how this number describes how the temperature is changing.
5. Determine the average rate of change of \(H^{-1}\) on the interval \([15,20]\text{.}\) Write a complete sentence to explain the meaning of the value you find, including units on the value. Explain clearly how this number describes how the number of chirps per minute is changing.

<1> Observe that \(g(f(t)) = g(2t+1) = \frac{(2t+1)-1}{2} = \frac{2t}{2} = t\text{.}\) Similarly, \(f(g(y)) = f\left(\frac{y-1}{2}\right) = 2\left(\frac{y-1}{2} \right) + 1 = y-1 + 1 = y\text{.}\)

<2> Calculus provides one way to fully justify that the graph of \(s\) is indeed always decreasing.