Inverses and Radical Functions

By the end of this section, you will be able to:

Find the inverse of a polynomial function
Restrict the domain to find the inverse of a polynomial function

Park rangers and other trail managers may construct rock piles, stacks, or other arrangements, usually called cairns, to mark trails or other landmarks. (Rangers and environmental scientists discourage hikers from doing the same, in order to avoid confusion and preserve the habitats of plants and animals.) A cairn in the form of a mound of gravel is in the shape of a cone with the height equal to twice the radius.

The volume is found using a formula from elementary geometry.

\begin{array}{lrcl} & V &=& \tfrac{1}{3}\pi r^2h \\[4pt] & &=& \tfrac{1}{3}\pi r^2(2r) \\[4pt] & &=& \tfrac{2}{3}\pi r^3 \end{array}

We have written the volume $V$ in terms of the radius $r$ . However, in some cases, we may start out with the volume and want to find the radius. For example: a customer purchases 100 cubic feet of gravel to construct a cone-shaped mound with a height twice the radius. What are the radius and height of the new cone? To answer this question, we use the formula

r=\sqrt[3]{\tfrac{3V}{2\pi}}

This function is the inverse of the formula for $V$ in terms of $r$ .

In this section, we will explore the inverses of polynomial and rational functions and, in particular, the radical functions we encounter in the process.

Finding the inverse of a polynomial function

Two functions $f$ and $g$ are inverse functions if for every coordinate pair in $f$ , $(a,b)$ , there exists a corresponding coordinate pair in the inverse function, $g$ , $(b,a)$ . In other words, the coordinate pairs of the inverse functions have the input and output interchanged.

For a function to have an inverse, it must be one-to-one.

For example, suppose the Sustainability Club builds a water runoff collector in the shape of a parabolic trough, shown below. We can use the information in the figure to find the surface area of the water in the trough as a function of the depth of the water.

Figure: A schematic diagram of a parabolic trough. The front cross section is the parabola y = x squared over 2, spanning 12 inches across the top rim from (-6, 18) to (6, 18) down to a vertex at the bottom, with an 18-inch height leader at the left. The same curve, offset up and to the right, represents the far end of the trough 3 feet away, with the top rims and bottom ridge connected by straight edges.

Because it will be helpful to have an equation for the parabolic cross-sectional shape, we will impose a coordinate system at the cross section, with $x$ measured horizontally and $y$ measured vertically, with the origin at the vertex of the parabola. See the graph below.

From this we find an equation for the parabolic shape. We placed the origin at the vertex of the parabola, so we know the equation will have form $y(x)=ax^2$ . Our equation will need to pass through the point $(6,18)$ , from which we can solve for the stretch factor $a$ .

\begin{array}{lrcl} & 18 &=& a\cdot6^2 \\[4pt] & a &=& \tfrac{18}{36} \\[4pt] & &=& \tfrac{1}{2} \end{array}

Our parabolic cross section has the equation

y(x)=\tfrac{1}{2}x^2

We are interested in the surface area of the water, so we must determine the width at the top of the water as a function of the water depth. For any depth $y$ the width will be given by $2x$ , so we need to solve the equation above for $x$ and find the inverse function. However, notice that the original function is not one-to-one, and indeed, given any output there are two inputs that produce the same output, one positive and one negative.

To find an inverse, we can restrict our original function to a limited domain on which it is one-to-one. In this case, it makes sense to restrict ourselves to positive $x$ values. On this domain, we can find an inverse by solving for the input variable:

\begin{array}{lrcl} & y &=& \tfrac{1}{2}x^2 \\[4pt] & 2y &=& x^2 \\[4pt] & x &=& \pm\sqrt{2y} \\[4pt] \therefore & y &=& \pm\sqrt{2x} \end{array}

This is not a function as written. Since we are limiting ourselves to positive $x$ values in the original function, we can eliminate the negative solution, which gives us the inverse function we’re looking for.

y=\sqrt{2x}

Because $x$ is the distance from the center of the parabola to either side, the entire width of the water at the top will be $2x$ . The trough is 3 feet (36 inches) long, so the surface area will then be:

\begin{array}{lrcl} \text{Area} &=& l\cdot w \\[4pt] &=& 36\cdot2x \\[4pt] &=& 72x \\[4pt] &=& 72\sqrt{2y} \end{array}

This example illustrates two important points:

When finding the inverse of a quadratic, we have to limit ourselves to a domain on which the function is one-to-one.
The inverse of a quadratic function is a square root function. Both are toolkit functions and different types of power functions.

Functions involving roots are often called radical functions. While it is not possible to find an inverse of most polynomial functions, some basic polynomials do have inverses. Such functions are called invertible functions, and we use the notation $f^{-1}(x)$ .

Warning: $f^{-1}(x)$ is not the same as the reciprocal of the function $f(x)$ . This use of “ $-1$ ” is reserved to denote inverse functions. To denote the reciprocal of a function $f(x)$ , we would need to write $(f(x))^{-1}=\tfrac{1}{f(x)}$ .

An important relationship between inverse functions is that they “undo” each other. If $f^{-1}$ is the inverse of a function $f$ , then $f$ is the inverse of the function $f^{-1}$ . In other words, whatever the function $f$ does to $x$ , $f^{-1}$ undoes it — and vice versa. More formally, we write

f^{-1}(f(x))=x,\ \text{for all }x\text{ in the domain of }f

and

f(f^{-1}(x))=x,\ \text{for all }x\text{ in the domain of }f^{-1}

Verifying two functions are inverses of one another. Two functions, $f$ and $g$ , are inverses of one another if for all $x$ in the domain of $f$ and $g$ .

g(f(x))=f(g(x))=x

How to: given a polynomial function, find the inverse of the function by restricting the domain in such a way that the new function is one-to-one.

Replace $f(x)$ with $y$ .
Interchange $x$ and $y$ .
Solve for $y$ , and rename the function $f^{-1}(x)$ .

Example. Show that $f(x)=\tfrac{1}{x+1}$ and $f^{-1}(x)=\tfrac{1}{x}-1$ are inverses, for $x\ne0,-1$ .

Solution. We must show that $f^{-1}(f(x))=x$ and $f(f^{-1}(x))=x$ .

\begin{array}{lrcl} f^{-1}(f(x)) &=& f^{-1}\left(\tfrac{1}{x+1}\right) \\[4pt] &=& \cfrac{1}{\tfrac{1}{x+1}}-1 \\[4pt] &=& (x+1)-1 \\[4pt] &=& x \end{array}

\begin{array}{lrcl} f(f^{-1}(x)) &=& f\left(\tfrac{1}{x}-1\right) \\[4pt] &=& \cfrac{1}{\left(\tfrac{1}{x}-1\right)+1} \\[4pt] &=& \cfrac{1}{\tfrac{1}{x}} \\[4pt] &=& x \end{array}

Therefore, $f(x)=\tfrac{1}{x+1}$ and $f^{-1}(x)=\tfrac{1}{x}-1$ are inverses.

To begin verifying that $f(x)=\tfrac{x+5}{3}$ and $f^{-1}(x)=3x-5$ are inverses, evaluate $f(1)$ .

Now evaluate $f^{-1}$ at that result.

Example. Find the inverse of the function $f(x)=5x^3+1$ .

Solution. This is a transformation of the basic cubic toolkit function, and based on our knowledge of that function, we know it is one-to-one. Solving for the inverse by solving for $x$ .

\begin{array}{lrcl} & y &=& 5x^3+1 \\[4pt] & x &=& 5y^3+1 \\[4pt] & x-1 &=& 5y^3 \\[4pt] & \tfrac{x-1}{5} &=& y^3 \\[4pt] f^{-1}(x) &=& \sqrt[3]{\tfrac{x-1}{5}} \end{array}

Analysis. Look at the graph of $f$ and $f^{-1}$ below. Notice that the two graphs are symmetrical about the line $y=x$ . This is always the case when graphing a function and its inverse function.

Also, since the method involved interchanging $x$ and $y$ , notice corresponding points. If $(a,b)$ is on the graph of $f$ , then $(b,a)$ is on the graph of $f^{-1}$ . Since $(0,1)$ is on the graph of $f$ , then $(1,0)$ is on the graph of $f^{-1}$ . Similarly, since $(1,6)$ is on the graph of $f$ , then $(6,1)$ is on the graph of $f^{-1}$ .

Figure: The graph of f of x equals 5x cubed plus 1 in blue, rising steeply through (0, 1) and (1, 6), and its inverse f inverse of x equals the cube root of the quantity x minus 1, divided by 5, in red, passing through (1, 0) and (6, 1). A dashed line marks y = x, showing the two curves are mirror images across it.

Find the inverse function of $f(x)=\sqrt[3]{x+4}$ .

Restricting the domain to find the inverse of a polynomial function

So far, we have been able to find the inverse functions of cubic functions without having to restrict their domains. However, as we know, not all cubic polynomials are one-to-one. Some functions that are not one-to-one may have their domain restricted so that they are one-to-one, but only over that domain. The function over the restricted domain would then have an inverse function. Since quadratic functions are not one-to-one, we must restrict their domain in order to find their inverses.

Restricting the domain. If a function is not one-to-one, it cannot have an inverse. If we restrict the domain of the function so that it becomes one-to-one, thus creating a new function, this new function will have an inverse.

How to: given a polynomial function, restrict the domain of a function that is not one-to-one and then find the inverse.

Restrict the domain by determining a domain on which the original function is one-to-one.
Replace $f(x)$ with $y$ .
Interchange $x$ and $y$ .
Solve for $y$ , and rename the function or pair of function $f^{-1}(x)$ .
Revise the formula for $f^{-1}(x)$ by ensuring that the outputs of the inverse function correspond to the restricted domain of the original function.

Example. Find the inverse function of $f$ :

ⓐ $f(x)=(x-4)^2,\ x\ge4$

ⓑ $f(x)=(x-4)^2,\ x\le4$

Solution. The original function $f(x)=(x-4)^2$ is not one-to-one, but the function is restricted to a domain of $x\ge4$ or $x\le4$ on which it is one-to-one. See the graphs below.

Figure: The right half of an upward parabola with vertex at (4, 0), the graph of f of x equals the quantity x minus 4, squared, restricted to x greater than or equal to 4. The curve rises to the right from the vertex, with no arrowhead at the vertex since the domain begins there.

Figure: The left half of an upward parabola with vertex at (4, 0), the graph of f of x equals the quantity x minus 4, squared, restricted to x less than or equal to 4. The curve descends toward the vertex from the upper left, with no arrowhead at the vertex since the domain ends there.

To find the inverse, start by replacing $f(x)$ with the simple variable $y$ .

\begin{array}{lrcll} & y &=& (x-4)^2 & \text{Interchange }x\text{ and }y. \\[4pt] & x &=& (y-4)^2 & \text{Take the square root.} \\[4pt] & \pm\sqrt{x} &=& y-4 & \text{Add 4 to both sides.} \\[4pt] & 4\pm\sqrt{x} &=& y & \end{array}

This is not a function as written. We need to examine the restrictions on the domain of the original function to determine the inverse. Since we reversed the roles of $x$ and $y$ for the original $f(x)$ , we looked at the domain: the values $x$ could assume. When we reversed the roles of $x$ and $y$ , this gave us the values $y$ could assume. For this function, $x\ge4$ , so for the inverse, we should have $y\ge4$ , which is what our inverse function gives.

ⓐ The domain of the original function was restricted to $x\ge4$ , so the outputs of the inverse need to be the same, $f(x)\ge4$ , and we must use the $+$ case:

f^{-1}(x)=4+\sqrt{x}

ⓑ The domain of the original function was restricted to $x\le4$ , so the outputs of the inverse need to be the same, $f(x)\le4$ , and we must use the $-$ case:

f^{-1}(x)=4-\sqrt{x}

Analysis. On the graphs below, we see the original function graphed on the same set of axes as its inverse function. Notice that together the graphs show symmetry about the line $y=x$ . The coordinate pair $(4,0)$ is on the graph of $f$ and the coordinate pair $(0,4)$ is on the graph of $f^{-1}$ . For any coordinate pair, if $(a,b)$ is on the graph of $f$ , then $(b,a)$ is on the graph of $f^{-1}$ . Finally, observe that the graph of $f$ intersects the graph of $f^{-1}$ on the line $y=x$ . Points of intersection for the graphs of $f$ and $f^{-1}$ will always lie on the line $y=x$ .

Figure: The curve labelled f, the right half of an upward parabola with vertex at (4, 0) restricted to x greater than or equal to 4, together with its inverse, the curve labelled f inverse, equal to 4 plus the square root of x, rising from (0, 4). A dashed line marks y = x. The two curves are mirror images across it, and the points (4, 0) and (0, 4) are marked.

Figure: The curve labelled f, the left half of an upward parabola with vertex at (4, 0) restricted to x less than or equal to 4, together with its inverse, the curve labelled f inverse, equal to 4 minus the square root of x, descending from (0, 4). A dashed line marks y = x. The two curves are mirror images across it, and the points (4, 0) and (0, 4) are marked.

Example. Restrict the domain and then find the inverse of

f(x)=(x-2)^2-3.

Solution. We can see this is a parabola with vertex at $(2,-3)$ that opens upward. Because the graph will be decreasing on one side of the vertex and increasing on the other side, we can restrict this function to a domain on which it will be one-to-one by limiting the domain to $x\ge2$ .

To find the inverse, we will use the vertex form of the quadratic. We start by replacing $f(x)$ with a simple variable, $y$ , then solve for $x$ .

\begin{array}{lrcll} & y &=& (x-2)^2-3 & \text{Interchange }x\text{ and }y. \\[4pt] & x &=& (y-2)^2-3 & \text{Add 3 to both sides.} \\[4pt] & x+3 &=& (y-2)^2 & \text{Take the square root.} \\[4pt] & \pm\sqrt{x+3} &=& y-2 & \text{Add 2 to both sides.} \\[4pt] & 2\pm\sqrt{x+3} &=& y & \text{Rename the function.} \\[4pt] f^{-1}(x) &=& 2\pm\sqrt{x+3} & \end{array}

Now we need to determine which case to use. Because we restricted our original function to a domain of $x\ge2$ , the outputs of the inverse should be the same, telling us to utilize the $+$ case

f^{-1}(x)=2+\sqrt{x+3}

If the quadratic had not been given in vertex form, rewriting it into vertex form would be the first step. This way we may easily observe the coordinates of the vertex to help us restrict the domain.

Analysis. Notice that we arbitrarily decided to restrict the domain on $x\ge2$ . We could just have easily opted to restrict the domain on $x\le2$ , in which case $f^{-1}(x)=2-\sqrt{x+3}$ . Observe the original function graphed on the same set of axes as its inverse function below. Notice that both graphs show symmetry about the line $y=x$ . The coordinate pair $(2,-3)$ is on the graph of $f$ and the coordinate pair $(-3,2)$ is on the graph of $f^{-1}$ . Observe from the graph of both functions on the same set of axes that

\text{domain of }f=\text{range of }f^{-1}=[2,\infty)

and

\text{domain of }f^{-1}=\text{range of }f=[-3,\infty)

Finally, observe that the graph of $f$ intersects the graph of $f^{-1}$ along the line $y=x$ .

Figure: The curve labelled f, the right half of an upward parabola with vertex at (2, -3) restricted to x greater than or equal to 2, together with its inverse, the curve labelled f inverse, equal to 2 plus the square root of the quantity x plus 3, rising from (-3, 2). A dashed line marks y = x. The two curves are mirror images across it, and the points (2, -3) and (-3, 2) are marked.

Find the inverse of the function $f(x)=x^2+1$ , on the domain $x\ge0$ .

Solving applications of radical functions

Notice that the functions from previous examples were all polynomials, and their inverses were radical functions. If we want to find the inverse of a radical function, we will need to restrict the domain of the answer because the range of the original function is limited.

How to: given a radical function, find the inverse.

Determine the range of the original function.
Replace $f(x)$ with $y$ , then solve for $x$ .
If necessary, restrict the domain of the inverse function to the range of the original function.

Example. Restrict the domain and then find the inverse of the function $f(x)=\sqrt{x-4}$ .

Solution. Note that the original function has range $f(x)\ge0$ . Replace $f(x)$ with $y$ , then solve for $x$ .

\begin{array}{lrcl} y &=& \sqrt{x-4} & \text{Replace }f(x)\text{ with }y. \\[4pt] x &=& \sqrt{y-4} & \text{Interchange }x\text{ and }y. \\[4pt] x^2 &=& y-4 & \text{Square each side.} \\[4pt] x^2+4 &=& y & \text{Add 4.} \\[4pt] f^{-1}(x) &=& x^2+4 & \text{Rename the function }f^{-1}(x). \end{array}

Recall that the domain of this function must be limited to the range of the original function.

f^{-1}(x)=x^2+4,\ x\ge0

Analysis. Notice below that the inverse is a reflection of the original function over the line $y=x$ . Because the original function has only positive outputs, the inverse function has only positive inputs.

Figure: The curve labelled f, the square root of the quantity x minus 4, restricted to x greater than or equal to 4, together with its inverse, the curve labelled f inverse, equal to x squared plus 4, restricted to x greater than or equal to 0. Neither curve has an arrowhead at its starting endpoint, since each domain begins there. A dashed line marks y = x, and the two curves are mirror images across it.

Restrict the domain and then find the inverse of the function $f(x)=\sqrt{2x+3}$ .

Radical functions are common in physical models, as we saw in the section opener. We now have enough tools to be able to solve the problem posed at the start of the section.

Example. Park rangers construct a mound of gravel in the shape of a cone with the height equal to twice the radius. The volume of the cone in terms of the radius is given by

V=\tfrac{2}{3}\pi r^3

Find the inverse of the function $V=\tfrac{2}{3}\pi r^3$ that determines the volume $V$ of a cone and is a function of the radius $r$ . Then use the inverse function to calculate the radius of such a mound of gravel measuring 100 cubic feet. Use $\pi=3.14$ .

Solution. Start with the given function for $V$ . Notice that the meaningful domain for the function is $r\ge0$ since negative radii would not make sense in this context. Also note the range of the function (hence, the domain of the inverse function) is $V\ge0$ . Solve for $r$ in terms of $V$ , using the method outlined previously.

\begin{array}{lrcll} & V &=& \tfrac{2}{3}\pi r^3 & \\[4pt] & r^3 &=& \tfrac{3V}{2\pi} & \text{Solve for }r^3. \\[4pt] & r &=& \sqrt[3]{\tfrac{3V}{2\pi}} & \text{Solve for }r. \end{array}

This is the result stated in the section opener. Now evaluate this for $V=100$ and $\pi=3.14$ .

\begin{array}{lrcl} r &=& \sqrt[3]{\tfrac{3V}{2\pi}} \\[4pt] &=& \sqrt[3]{\tfrac{3\cdot100}{2\cdot3.14}} \\[4pt] &\approx& \sqrt[3]{47.7707} \\[4pt] &\approx& 3.63 \end{array}

Therefore, the radius is about 3.63 ft.

Determining the domain of a radical function composed with other functions

When radical functions are composed with other functions, determining domain can become more complicated.

Example. Find the domain of the function $f(x)=\sqrt{\tfrac{(x+2)(x-3)}{(x-1)}}$ .

Solution. Because a square root is only defined when the quantity under the radical is non-negative, we need to determine where $\tfrac{(x+2)(x-3)}{(x-1)}\ge0$ . The output of a rational function can change signs (change from positive to negative or vice versa) at $x$ -intercepts and at vertical asymptotes. For this equation, the graph could change signs at $x=-2$ , $1$ , and $3$ .

To determine the intervals on which the rational expression is positive, we could test some values in the expression or sketch a graph. While both approaches work equally well, for this example we will use a graph, shown below.

Figure: The graph of the rational expression (x plus 2)(x minus 3) over (x minus 1), with x-intercepts at (-2, 0) and (3, 0) and a vertical asymptote at x = 1, dashed. The curve dips below the x-axis between -2 and 1, and again is negative approaching the asymptote from the right before crossing at x = 3. The outputs are non-negative — the curve is on or above the x-axis — on the intervals from -2 up to 1 and from 3 onward, each marked Outputs are non-negative.

This function has two $x$ -intercepts, both of which exhibit linear behavior near the $x$ -intercepts. There is one vertical asymptote, corresponding to a linear factor; this behavior is similar to the basic reciprocal toolkit function, and there is no horizontal asymptote because the degree of the numerator is larger than the degree of the denominator. There is a $y$ -intercept at $(0,\sqrt6)$ .

From the $y$ -intercept and $x$ -intercept at $x=-2$ , we can sketch the left side of the graph. From the behavior at the asymptote, we can sketch the right side of the graph.

From the graph, we can now tell on which intervals the outputs will be non-negative, so that we can be sure that the original function $f(x)$ will be defined. $f(x)$ has domain $-2\le x<1$ or $x\ge3$ , or in interval notation, $[-2,1)\cup[3,\infty)$ .

Finding inverses of rational functions

As with finding inverses of quadratic functions, it is sometimes desirable to find the inverse of a rational function, particularly of rational functions that are the ratio of linear functions, such as in concentration applications.

Example. The function $C=\tfrac{20+0.4n}{100+n}$ represents the concentration $C$ of an acid solution after $n$ mL of 40% solution has been added to 100 mL of a 20% solution. First, find the inverse of the function; that is, find an expression for $n$ in terms of $C$ . Then use your result to determine how much of the 40% solution should be added so that the final mixture is a 35% solution.

Solution. We first want the inverse of the function. We will solve for $n$ in terms of $C$ .

\begin{array}{lrcl} & C &=& \tfrac{20+0.4n}{100+n} \\[4pt] & C(100+n) &=& 20+0.4n \\[4pt] & 100C+Cn &=& 20+0.4n \\[4pt] & 100C-20 &=& 0.4n-Cn \\[4pt] & 100C-20 &=& (0.4-C)n \\[4pt] & n &=& \tfrac{100C-20}{0.4-C} \end{array}

Now evaluate this function for $C=0.35$ (35%).

\begin{array}{lrcl} n &=& \tfrac{100(0.35)-20}{0.4-0.35} \\[4pt] &=& \tfrac{15}{0.05} \\[4pt] &=& 300 \end{array}

We can conclude that 300 mL of the 40% solution should be added.

Find the inverse of the function $f(x)=\tfrac{x+3}{x-2}$ .

Key concepts

The inverse of a quadratic function is a square root function.
If $f^{-1}$ is the inverse of a function $f$ , then $f$ is the inverse of the function $f^{-1}$ .
While it is not possible to find an inverse of most polynomial functions, some basic polynomials are invertible.
To find the inverse of certain functions, we must restrict the function to a domain on which it will be one-to-one.
When finding the inverse of a radical function, we need a restriction on the domain of the answer.
Inverse and radical functions can be used to solve application problems.

Practice

Find the inverse of a polynomial function

Find the inverse of the function $f(x)=x^3+5$ .

Find the inverse of the function $f(x)=\tfrac{3x+4}{5-4x}$ .

The volume, $V$ , of a sphere in terms of its radius, $r$ , is given by $V(r)=\tfrac{4}{3}\pi r^3$ . Express $r$ as a function of $V$ .

Using that same inverse, find the radius of a sphere with a volume of 200 cubic feet, rounded to the nearest hundredth.

Restrict the domain to find the inverse of a polynomial function

Find the inverse of the function $f(x)=(x+1)^2-3$ on the domain $x\ge-1$ .

Find the inverse of the function $f(x)=3x^2+5$ on the domain $x\le0$ .

The volume of a right circular cone, $V$ , in terms of its radius, $r$ , and a fixed height of 12 inches, is given by $V=4\pi r^2$ . Restricting to the physically meaningful domain $r\ge0$ , express $r$ as a function of $V$ .

Using that same inverse, find the radius of a cone with volume 50 cubic inches, rounded to the nearest hundredth.

This section is adapted from Precalculus 2e, Section 3.8: Inverses and Radical Functions by Jay Abramson and OpenStax, © OpenStax, licensed under CC BY-NC-SA 4.0. Access the original for free at openstax.org. Changes: recreated every figure as an accessible inline SVG generated from its exact equation — the parabolic-trough setup diagram as a labelled schematic drawn from its own cross-sectional formula $y=\tfrac{1}{2}x^2$ , the standalone parabola $y=\tfrac{1}{2}x^2$ , the cubic $f(x)=5x^3+1$ with its cube-root inverse and the line $y=x$ , the two half-parabola panels for $f(x)=(x-4)^2$ restricted to $x\ge4$ and $x\le4$ , the matching two panels pairing each half with its square-root inverse, the quadratic $f(x)=(x-2)^2-3$ restricted to $x\ge2$ with its inverse $f^{-1}(x)=2+\sqrt{x+3}$ , the radical $f(x)=\sqrt{x-4}$ with its quadratic inverse, and the rational sign-analysis graph of $\tfrac{(x+2)(x-3)}{(x-1)}$ ; omitted the decorative gravel-cone photograph, which carries no mathematics; converted the five “Try It” checks into interactive fill-in components, including rephrasing the two verify-that-two-functions-are-inverses checks (Example 1’s Try It) as a numerical composition check, $f^{-1}(f(1))$ , since a free-response box cannot grade a written proof; and adapted six selected end-of-section exercises — a cubic-function inverse and a rational-function inverse, two quadratic inverses on a source-given restricted domain, and two real-world cone/sphere volume-inverse problems, each split into a formula-then-evaluate pair of prompts since it already asks for both a general inverse and a numerical evaluation — into eight interactive components in a closing Practice block, one group per objective.

Rational Functions Modeling Using Variation