23
Learning Objectives
In this section, you will:
- Verify inverse functions.
- Determine the domain and range of an inverse function, and restrict the domain of a function to make it one-to-one.
- Find or evaluate the inverse of a function.
- Use the graph of a one-to-one function to graph its inverse function on the same axes.
A reversible heat pump is a climate-control system that is an air conditioner and a heater in a single device. Operated in one direction, it pumps heat out of a house to provide cooling. Operating in reverse, it pumps heat into the building from the outside, even in cool weather, to provide heating. As a heater, a heat pump is several times more efficient than conventional electrical resistance heating.
If some physical machines can run in two directions, we might ask whether some of the function “machines” we have been studying can also run backwards. (Figure) provides a visual representation of this question. In this section, we will consider the reverse nature of functions.
Verifying That Two Functions Are Inverse Functions
Suppose a fashion designer traveling to Milan for a fashion show wants to know what the temperature will be. He is not familiar with the Celsius scale. To get an idea of how temperature measurements are related, he asks his assistant, Betty, to convert 75 degrees Fahrenheit to degrees Celsius. She finds the formula
and substitutes 75 for[latex]\,F\,[/latex]to calculate
Knowing that a comfortable 75 degrees Fahrenheit is about 24 degrees Celsius, he sends his assistant the week’s weather forecast from (Figure) for Milan, and asks her to convert all of the temperatures to degrees Fahrenheit.
At first, Betty considers using the formula she has already found to complete the conversions. After all, she knows her algebra, and can easily solve the equation for[latex]\,F\,[/latex]after substituting a value for[latex]\,C.\,[/latex]For example, to convert 26 degrees Celsius, she could write
After considering this option for a moment, however, she realizes that solving the equation for each of the temperatures will be awfully tedious. She realizes that since evaluation is easier than solving, it would be much more convenient to have a different formula, one that takes the Celsius temperature and outputs the Fahrenheit temperature.
The formula for which Betty is searching corresponds to the idea of an inverse function, which is a function for which the input of the original function becomes the output of the inverse function and the output of the original function becomes the input of the inverse function.
Given a function[latex]\,f\left(x\right),\,[/latex]we represent its inverse as[latex]\,{f}^{-1}\left(x\right),\,[/latex]read as[latex]\,“f\,[/latex]inverse of[latex]\,x.\text{”}\,[/latex]The raised[latex]\,-1\,[/latex]is part of the notation. It is not an exponent; it does not imply a power of[latex]\,-1\,[/latex]. In other words,[latex]\,{f}^{-1}\left(x\right)\,[/latex]does not mean[latex]\,\frac{1}{f\left(x\right)}\,[/latex]because[latex]\,\frac{1}{f\left(x\right)}\,[/latex]is the reciprocal of[latex]\,f\,[/latex]and not the inverse.
The “exponent-like” notation comes from an analogy between function composition and multiplication: just as[latex]\,{a}^{-1}a=1\,[/latex](1 is the identity element for multiplication) for any nonzero number[latex]\,a,\,[/latex]so[latex]\,{f}^{-1}\circ f\,[/latex]equals the identity function, that is,
This holds for all[latex]\,x\,[/latex]in the domain of[latex]\,f.\,[/latex]Informally, this means that inverse functions “undo” each other. However, just as zero does not have a reciprocal, some functions do not have inverses.
Given a function[latex]\,f\left(x\right),\,[/latex]we can verify whether some other function[latex]\,g\left(x\right)\,[/latex]is the inverse of[latex]\,f\left(x\right)\,[/latex]by checking whether either[latex]\,g\left(f\left(x\right)\right)=x\,[/latex]or[latex]\,f\left(g\left(x\right)\right)=x\,[/latex]is true. We can test whichever equation is more convenient to work with because they are logically equivalent (that is, if one is true, then so is the other.)
For example,[latex]\,y=4x\,[/latex]and[latex]\,y=\frac{1}{4}x\,[/latex]are inverse functions.
and
A few coordinate pairs from the graph of the function[latex]\,y=4x\,[/latex]are (−2, −8), (0, 0), and (2, 8). A few coordinate pairs from the graph of the function[latex]\,y=\frac{1}{4}x\,[/latex]are (−8, −2), (0, 0), and (8, 2). If we interchange the input and output of each coordinate pair of a function, the interchanged coordinate pairs would appear on the graph of the inverse function.
Inverse Function
For any one-to-one function[latex]\,f\left(x\right)=y,\,[/latex]a function[latex]\,{f}^{-1}\left(x\right)\,[/latex]is an inverse function of[latex]\,f\,[/latex]if[latex]\,{f}^{-1}\left(y\right)=x.\,[/latex]This can also be written as[latex]\,{f}^{-1}\left(f\left(x\right)\right)=x\,[/latex]for all[latex]\,x\,[/latex]in the domain of[latex]\,f.\,[/latex]It also follows that[latex]\,f\left({f}^{-1}\left(x\right)\right)=x\,[/latex]for all[latex]\,x\,[/latex]in the domain of[latex]\,{f}^{-1}\,[/latex]if[latex]\,{f}^{-1}\,[/latex]is the inverse of[latex]\,f.\,[/latex]
The notation [latex]{f}^{-1}[/latex] is read [latex]\text{“}f[/latex] inverse.” Like any other function, we can use any variable name as the input for [latex]{f}^{-1},[/latex] so we will often write[latex]\,{f}^{-1}\left(x\right),[/latex] which we read as [latex]“f[/latex] inverse of [latex]x.”[/latex]
Keep in mind that
and not all functions have inverses.
Identifying an Inverse Function for a Given Input-Output Pair
If for a particular one-to-one function[latex]\,f\left(2\right)=4\,[/latex]and[latex]\,f\left(5\right)=12,\,[/latex]what are the corresponding input and output values for the inverse function?
Show Solution
The inverse function reverses the input and output quantities, so if
Alternatively, if we want to name the inverse function[latex]\,g,\,[/latex]then[latex]\,g\left(4\right)=2\,[/latex]and[latex]\,g\left(12\right)=5.[/latex]
Analysis
Notice that if we show the coordinate pairs in a table form, the input and output are clearly reversed. See (Figure).
| [latex]\left(x,f\left(x\right)\right)[/latex] | [latex]\left(x,g\left(x\right)\right)[/latex] |
|---|---|
| [latex]\left(2,4\right)[/latex] | [latex]\left(4,2\right)[/latex] |
| [latex]\left(5,12\right)[/latex] | [latex]\left(12,5\right)[/latex] |
Try It
Given that[latex]\,{h}^{-1}\left(6\right)=2,\,[/latex]what are the corresponding input and output values of the original function[latex]\,h?\,[/latex]
Show Solution
[latex]h\left(2\right)=6[/latex]
How To
Given two functions[latex]\,\,f\left(x\right)\,\,[/latex]and[latex]\,g\left(x\right),\,[/latex]test whether the functions are inverses of each other.
- Determine whether[latex]\,f\left(g\left(x\right)\right)=x\,[/latex]or[latex]\,g\left(f\left(x\right)\right)=x.[/latex]
- If either statement is true, then both are true, and[latex]\,g={f}^{-1}\,[/latex]and[latex]\,f={g}^{-1}.\,[/latex]If either statement is false, then both are false, and[latex]\,g\ne {f}^{-1}\,[/latex]and[latex]\,f\ne {g}^{-1}.[/latex]
Testing Inverse Relationships Algebraically
If[latex]\,f\left(x\right)=\frac{1}{x+2}\,[/latex]and[latex]\,g\left(x\right)=\frac{1}{x}-2,\,[/latex]is[latex]\,g={f}^{-1}?[/latex]
Show Solution
so
This is enough to answer yes to the question, but we can also verify the other formula.
Analysis
Notice the inverse operations are in reverse order of the operations from the original function.
Try It
If[latex]\,f\left(x\right)={x}^{3}-4\,[/latex]and[latex]\,g\left(x\right)=\sqrt[\,3]{x+4},\,[/latex]is[latex]\,g={f}^{-1}?[/latex]
Show Solution
Yes
Determining Inverse Relationships for Power Functions
If[latex]\,f\left(x\right)={x}^{3}\,[/latex](the cube function) and[latex]\,g\left(x\right)=\frac{1}{3}x,\,[/latex]is[latex]\,g={f}^{-1}?[/latex]
Show Solution
No, the functions are not inverses.
Analysis
The correct inverse to the cube is, of course, the cube root[latex]\,\sqrt[3]{x}={x}^{\frac{1}{3}},\,[/latex]that is, the one-third is an exponent, not a multiplier.
Try It
If[latex]\,f\left(x\right)={\left(x-1\right)}^{3}\,\text{and}\,g\left(x\right)=\sqrt[3]{x}+1,\,[/latex]is[latex]\,g={f}^{-1}?[/latex]
Show Solution
Yes
Finding Domain and Range of Inverse Functions
The outputs of the function[latex]\,f\,[/latex]are the inputs to[latex]\,{f}^{-1},\,[/latex]so the range of[latex]\,f\,[/latex]is also the domain of[latex]\,{f}^{-1}.\,[/latex]Likewise, because the inputs to[latex]\,f\,[/latex]are the outputs of[latex]\,{f}^{-1},\,[/latex]the domain of[latex]\,f\,[/latex]is the range of[latex]\,{f}^{-1}.\,[/latex]We can visualize the situation as in (Figure).
When a function has no inverse function, it is possible to create a new function where that new function on a limited domain does have an inverse function. For example, the inverse of[latex]\,f\left(x\right)=\sqrt{x}\,[/latex]is[latex]\,{f}^{-1}\left(x\right)={x}^{2},\,[/latex]because a square “undoes” a square root; but the square is only the inverse of the square root on the domain[latex]\,\left[0,\infty \right),\,[/latex]since that is the range of[latex]\,f\left(x\right)=\sqrt{x}.[/latex]
We can look at this problem from the other side, starting with the square (toolkit quadratic) function[latex]\,f\left(x\right)={x}^{2}.\,[/latex]If we want to construct an inverse to this function, we run into a problem, because for every given output of the quadratic function, there are two corresponding inputs (except when the input is 0). For example, the output 9 from the quadratic function corresponds to the inputs 3 and –3. But an output from a function is an input to its inverse; if this inverse input corresponds to more than one inverse output (input of the original function), then the “inverse” is not a function at all! To put it differently, the quadratic function is not a one-to-one function; it fails the horizontal line test, so it does not have an inverse function. In order for a function to have an inverse, it must be a one-to-one function.
In many cases, if a function is not one-to-one, we can still restrict the function to a part of its domain on which it is one-to-one. For example, we can make a restricted version of the square function[latex]\,f\left(x\right)={x}^{2}\,[/latex]with its domain limited to[latex]\,\left[0,\infty \right),\,[/latex]which is a one-to-one function (it passes the horizontal line test) and which has an inverse (the square-root function).
If[latex]\,f\left(x\right)={\left(x-1\right)}^{2}\,[/latex]on[latex]\,\left[1,\infty \right),\,[/latex]then the inverse function is[latex]\,{f}^{-1}\left(x\right)=\sqrt{x}+1.[/latex]
- The domain of[latex]\,f\,[/latex]= range of[latex]\,{f}^{-1}\,[/latex]=[latex]\,\left[1,\infty \right).[/latex]
- The domain of[latex]\,{f}^{-1}\,[/latex]= range of[latex]\,f\,[/latex]=[latex]\,\left[0,\infty \right).[/latex]
Is it possible for a function to have more than one inverse?
No. If two supposedly different functions, say,[latex]\,g\,[/latex]and[latex]\,h,\,[/latex]both meet the definition of being inverses of another function[latex]\,f,\,[/latex]then you can prove that[latex]\,g=h.\,[/latex]We have just seen that some functions only have inverses if we restrict the domain of the original function. In these cases, there may be more than one way to restrict the domain, leading to different inverses. However, on any one domain, the original function still has only one unique inverse.
Domain and Range of Inverse Functions
The range of a function[latex]\,f\left(x\right)\,[/latex]is the domain of the inverse function[latex]\,{f}^{-1}\left(x\right).[/latex]
The domain of[latex]\,f\left(x\right)\,[/latex]is the range of[latex]\,{f}^{-1}\left(x\right).[/latex]
How To
Given a function, find the domain and range of its inverse.
- If the function is one-to-one, write the range of the original function as the domain of the inverse, and write the domain of the original function as the range of the inverse.
- If the domain of the original function needs to be restricted to make it one-to-one, then this restricted domain becomes the range of the inverse function.
Finding the Inverses of Toolkit Functions
Identify which of the toolkit functions besides the quadratic function are not one-to-one, and find a restricted domain on which each function is one-to-one, if any. The toolkit functions are reviewed in (Figure). We restrict the domain in such a fashion that the function assumes all y-values exactly once.
| Constant | Identity | Quadratic | Cubic | Reciprocal |
|---|---|---|---|---|
| [latex]f\left(x\right)=c[/latex] | [latex]f\left(x\right)=x[/latex] | [latex]f\left(x\right)={x}^{2}[/latex] | [latex]f\left(x\right)={x}^{3}[/latex] | [latex]f\left(x\right)=\frac{1}{x}[/latex] |
| Reciprocal squared | Cube root | Square root | Absolute value | |
| [latex]f\left(x\right)=\frac{1}{{x}^{2}}[/latex] | [latex]f\left(x\right)=\sqrt[3]{x}[/latex] | [latex]f\left(x\right)=\sqrt{x}[/latex] | [latex]f\left(x\right)=|x|[/latex] |
Show Solution
The constant function is not one-to-one, and there is no domain (except a single point) on which it could be one-to-one, so the constant function has no inverse.
The absolute value function can be restricted to the domain[latex]\,\left[0,\infty \right),[/latex]where it is equal to the identity function.
The reciprocal-squared function can be restricted to the domain[latex]\,\left(0,\infty \right).[/latex]
Analysis
We can see that these functions (if unrestricted) are not one-to-one by looking at their graphs, shown in (Figure). They both would fail the horizontal line test. However, if a function is restricted to a certain domain so that it passes the horizontal line test, then in that restricted domain, it can have an inverse.
Try It
The domain of function[latex]\,f\,[/latex]is[latex]\,\left(1,\infty \right)\,[/latex]and the range of function[latex]\,f\,[/latex]is[latex]\,\left(\mathrm{-\infty },-2\right).\,[/latex]Find the domain and range of the inverse function.
Show Solution
The domain of function[latex]\,{f}^{-1}\,[/latex]is[latex]\,\left(-\infty \text{,}-2\right)\,[/latex]and the range of function[latex]\,{f}^{-1}\,[/latex]is[latex]\,\left(1,\infty \right).[/latex]
Finding and Evaluating Inverse Functions
Once we have a one-to-one function, we can evaluate its inverse at specific inverse function inputs or construct a complete representation of the inverse function in many cases.
Inverting Tabular Functions
Suppose we want to find the inverse of a function represented in table form. Remember that the domain of a function is the range of the inverse and the range of the function is the domain of the inverse. So we need to interchange the domain and range.
Each row (or column) of inputs becomes the row (or column) of outputs for the inverse function. Similarly, each row (or column) of outputs becomes the row (or column) of inputs for the inverse function.
Interpreting the Inverse of a Tabular Function
A function[latex]\,f\left(t\right)\,[/latex]is given in (Figure), showing distance in miles that a car has traveled in[latex]\,t\,[/latex]minutes. Find and interpret[latex]\,{f}^{-1}\left(70\right).[/latex]
| [latex]t\text{ (minutes)}[/latex] | 30 | 50 | 70 | 90 |
| [latex]f\left(t\right)\text{ (miles)}[/latex] | 20 | 40 | 60 | 70 |
Show Solution
The inverse function takes an output of[latex]\,f\,[/latex]and returns an input for[latex]\,f.\,[/latex]So in the expression[latex]\,{f}^{-1}\left(70\right),\,[/latex]70 is an output value of the original function, representing 70 miles. The inverse will return the corresponding input of the original function[latex]\,f,\,[/latex]90 minutes, so[latex]\,{f}^{-1}\left(70\right)=90.\,[/latex]The interpretation of this is that, to drive 70 miles, it took 90 minutes.
Alternatively, recall that the definition of the inverse was that if[latex]\,f\left(a\right)=b,\,[/latex]then[latex]\,{f}^{-1}\left(b\right)=a.\,[/latex]By this definition, if we are given[latex]\,{f}^{-1}\left(70\right)=a,\,[/latex]then we are looking for a value[latex]\,a\,[/latex]so that[latex]\,f\left(a\right)=70.\,[/latex]In this case, we are looking for a[latex]\,t\,[/latex]so that[latex]\,f\left(t\right)=70,\,[/latex]which is when[latex]\,t=90.[/latex]
Try It
Using (Figure), find and interpret (a)[latex]\text{ }f\left(60\right),[/latex]and (b)[latex]\text{ }{f}^{-1}\left(60\right).[/latex]
| [latex]t\text{ (minutes)}[/latex] | 30 | 50 | 60 | 70 | 90 |
| [latex]f\left(t\right)\text{ (miles)}[/latex] | 20 | 40 | 50 | 60 | 70 |
Show Solution
- [latex]f\left(60\right)=50.\,[/latex]In 60 minutes, 50 miles are traveled.
- [latex]{f}^{-1}\left(60\right)=70.\,[/latex]To travel 60 miles, it will take 70 minutes.
Evaluating the Inverse of a Function, Given a Graph of the Original Function
We saw in Functions and Function Notation that the domain of a function can be read by observing the horizontal extent of its graph. We find the domain of the inverse function by observing the vertical extent of the graph of the original function, because this corresponds to the horizontal extent of the inverse function. Similarly, we find the range of the inverse function by observing the horizontal extent of the graph of the original function, as this is the vertical extent of the inverse function. If we want to evaluate an inverse function, we find its input within its domain, which is all or part of the vertical axis of the original function’s graph.
How To
Given the graph of a function, evaluate its inverse at specific points.
- Find the desired input on the y-axis of the given graph.
- Read the inverse function’s output from the x-axis of the given graph.
Evaluating a Function and Its Inverse from a Graph at Specific Points
A function[latex]\,g\left(x\right)\,[/latex]is given in (Figure). Find[latex]\,g\left(3\right)\,[/latex]and[latex]\,{g}^{-1}\left(3\right).[/latex]
Show Solution
To evaluate [latex]g\left(3\right),\,[/latex]we find 3 on the x-axis and find the corresponding output value on the y-axis. The point [latex]\,\left(3,1\right)\,[/latex]tells us that[latex]\,g\left(3\right)=1.[/latex]
To evaluate[latex]\,{g}^{-1}\left(3\right),\,[/latex]recall that by definition[latex]\,{g}^{-1}\left(3\right)\,[/latex]means the value of x for which[latex]\,g\left(x\right)=3.\,[/latex]By looking for the output value 3 on the vertical axis, we find the point[latex]\,\left(5,3\right)\,[/latex]on the graph, which means[latex]\,g\left(5\right)=3,\,[/latex]so by definition,[latex]\,{g}^{-1}\left(3\right)=5.\,[/latex]See (Figure).
Try It
Using the graph in (Figure), (a) find[latex]\,{g}^{-1}\left(1\right),[/latex]and (b) estimate[latex]\,{g}^{-1}\left(4\right).[/latex]
Show Solution
a. 3; b. 5.6
Finding Inverses of Functions Represented by Formulas
Sometimes we will need to know an inverse function for all elements of its domain, not just a few. If the original function is given as a formula—for example,[latex]\,y\,[/latex]as a function of[latex]\,x\text{—}[/latex]we can often find the inverse function by solving to obtain[latex]\,x\,[/latex]as a function of[latex]\,y.[/latex]
How To
Given a function represented by a formula, find the inverse.
- Make sure[latex]\,f\,[/latex]is a one-to-one function.
- Solve for[latex]\,x.[/latex]
- Interchange[latex]\,x\,[/latex]and[latex]\,y.[/latex]
Inverting the Fahrenheit-to-Celsius Function
Find a formula for the inverse function that gives Fahrenheit temperature as a function of Celsius temperature.
Show Solution
By solving in general, we have uncovered the inverse function. If
then
In this case, we introduced a function[latex]\,h\,[/latex]to represent the conversion because the input and output variables are descriptive, and writing[latex]\,{C}^{-1}\,[/latex]could get confusing.
Try It
Solve for[latex]\,x\,[/latex]in terms of[latex]\,y\,[/latex]given[latex]\,y=\frac{1}{3}\left(x-5\right).[/latex]
Show Solution
[latex]x=3y+5[/latex]
Solving to Find an Inverse Function
Find the inverse of the function[latex]\,f\left(x\right)=\frac{2}{x-3}+4.[/latex]
Show Solution
So[latex]\,{f}^{-1}\left(y\right)=\frac{2}{y-4}+3\,[/latex]or[latex]\,{f}^{-1}\left(x\right)=\frac{2}{x-4}+3.[/latex]
Analysis
The domain and range of[latex]\,f\,[/latex]exclude the values 3 and 4, respectively.[latex]\,f\,[/latex] and [latex]\,{f}^{-1}\,[/latex]are equal at two points but are not the same function, as we can see by creating (Figure).
| [latex]x[/latex] | 1 | 2 | 5 | [latex]{f}^{-1}\left(y\right)[/latex] |
| [latex]f\left(x\right)[/latex] | 3 | 2 | 5 | [latex]y[/latex] |
Solving to Find an Inverse with Radicals
Find the inverse of the function[latex]\,f\left(x\right)=2+\sqrt{x-4}.[/latex]
Show Solution
So[latex]\,{f}^{-1}\left(x\right)={\left(x-2\right)}^{2}+4.[/latex]
The domain of[latex]\,f\,[/latex]is[latex]\,\left[4,\infty \right).\,[/latex]Notice that the range of[latex]\,f\,[/latex]is[latex]\,\left[2,\infty \right),\,[/latex]so this means that the domain of the inverse function[latex]\,{f}^{-1}\,[/latex]is also[latex]\,\left[2,\infty \right).[/latex]
Analysis
The formula we found for[latex]\,{f}^{-1}\left(x\right)\,[/latex]looks like it would be valid for all real[latex]\,x.\,[/latex]However,[latex]\,{f}^{-1}\,[/latex]itself must have an inverse (namely,[latex]\,f\,[/latex]) so we have to restrict the domain of[latex]\,{f}^{-1}\,[/latex]to[latex]\,\left[2,\infty \right)\,[/latex]in order to make[latex]\,{f}^{-1}\,[/latex]a one-to-one function. This domain of[latex]\,{f}^{-1}\,[/latex]is exactly the range of[latex]\,f.[/latex]
Try It
What is the inverse of the function[latex]\,f\left(x\right)=2-\sqrt{x}?[/latex]State the domains of both the function and the inverse function.
Show Solution
[latex]{f}^{-1}\left(x\right)={\left(2-x\right)}^{2};\,\,\text{domain}\,\,\text{of}\,\,f:\,\,\left[0,\infty \right);\,\,\text{domain}\,\,\text{of}\,\,{f}^{-1}:\,\,\left(-\infty ,2\right][/latex]
Finding Inverse Functions and Their Graphs
Now that we can find the inverse of a function, we will explore the graphs of functions and their inverses. Let us return to the quadratic function[latex]\,f\left(x\right)={x}^{2}\,[/latex]restricted to the domain[latex]\,\left[0,\infty \right)\text{,}[/latex] on which this function is one-to-one, and graph it as in (Figure).
Restricting the domain to[latex]\,\left[0,\infty \right)\,[/latex]makes the function one-to-one (it will obviously pass the horizontal line test), so it has an inverse on this restricted domain.
We already know that the inverse of the toolkit quadratic function is the square root function, that is, [latex]{f}^{-1}\left(x\right)=\sqrt{x}.[/latex] What happens if we graph both [latex]f\text{ }[/latex] and [latex]{f}^{-1}[/latex] on the same set of axes, using the [latex]x\text{-}[/latex]axis for the input to both [latex]f\text{ and }{f}^{-1}?[/latex]
We notice a distinct relationship: The graph of[latex]\,{f}^{-1}\left(x\right)\,[/latex]is the graph of[latex]\,f\left(x\right)\,[/latex]reflected about the diagonal line[latex]\,y=x,\,[/latex]which we will call the identity line, shown in (Figure).
This relationship will be observed for all one-to-one functions, because it is a result of the function and its inverse swapping inputs and outputs. This is equivalent to interchanging the roles of the vertical and horizontal axes.
Finding the Inverse of a Function Using Reflection about the Identity Line
Given the graph of[latex]\,f\left(x\right)\,[/latex]in (Figure), sketch a graph of[latex]\,{f}^{-1}\left(x\right).[/latex]
Show Solution
This is a one-to-one function, so we will be able to sketch an inverse. Note that the graph shown has an apparent domain of[latex]\,\left(0,\infty \right)\,[/latex]and range of[latex]\,\left(-\infty ,\infty \right),\,[/latex]so the inverse will have a domain of[latex]\,\left(-\infty ,\infty \right)\,[/latex]and range of[latex]\,\left(0,\infty \right).[/latex]
If we reflect this graph over the line[latex]\,y=x,\,[/latex]the point[latex]\,\left(1,0\right)\,[/latex]reflects to[latex]\,\left(0,1\right)\,[/latex]and the point[latex]\,\left(4,2\right)\,[/latex]reflects to[latex]\,\left(2,4\right).\,[/latex]Sketching the inverse on the same axes as the original graph gives (Figure).
Try It
Draw graphs of the functions[latex]\,f\text{ }[/latex]and[latex]\text{ }{f}^{-1}[/latex]from (Figure).
Show Solution
Is there any function that is equal to its own inverse?
Yes. If[latex]\,f={f}^{-1},\,[/latex]then[latex]\,f\left(f\left(x\right)\right)=x,\,[/latex]and we can think of several functions that have this property. The identity function does, and so does the reciprocal function, because
Any function[latex]\,f\left(x\right)=c-x,\,[/latex]where[latex]\,c\,[/latex]is a constant, is also equal to its own inverse.
Access these online resources for additional instruction and practice with inverse functions.
Visit this website for additional practice questions from Learningpod.
Key Concepts
- If[latex]\,g\left(x\right)\,[/latex]is the inverse of[latex]\,f\left(x\right),\,[/latex]then[latex]\,g\left(f\left(x\right)\right)=f\left(g\left(x\right)\right)=x.\,[/latex]See (Figure), (Figure), and (Figure).
- Only some of the toolkit functions have an inverse. See (Figure).
- For a function to have an inverse, it must be one-to-one (pass the horizontal line test).
- A function that is not one-to-one over its entire domain may be one-to-one on part of its domain.
- For a tabular function, exchange the input and output rows to obtain the inverse. See (Figure).
- The inverse of a function can be determined at specific points on its graph. See (Figure).
- To find the inverse of a formula, solve the equation[latex]\,y=f\left(x\right)\,[/latex]for[latex]\,x\,[/latex]as a function of[latex]\,y.\,[/latex]Then exchange the labels[latex]\,x\,[/latex]and[latex]\,\,y.\,\,[/latex]See (Figure), (Figure), and (Figure).
- The graph of an inverse function is the reflection of the graph of the original function across the line[latex]\,y=x.\,[/latex]See (Figure).
Section Exercises
Verbal
Describe why the horizontal line test is an effective way to determine whether a function is one-to-one?
Show Solution
Each output of a function must have exactly one output for the function to be one-to-one. If any horizontal line crosses the graph of a function more than once, that means that[latex]\,y[/latex]-values repeat and the function is not one-to-one. If no horizontal line crosses the graph of the function more than once, then no[latex]\,y[/latex]-values repeat and the function is one-to-one.
Why do we restrict the domain of the function[latex]\,f\left(x\right)={x}^{2}\,[/latex]to find the function’s inverse?
Can a function be its own inverse? Explain.
Show Solution
Yes. For example,[latex]\,f\left(x\right)=\frac{1}{x}\,[/latex]is its own inverse.
Are one-to-one functions either always increasing or always decreasing? Why or why not?
How do you find the inverse of a function algebraically?
Show Solution
Given a function[latex]\,y=f\left(x\right),\,[/latex]solve for[latex]\,x\,[/latex]in terms of[latex]\,y.\,[/latex]Interchange the[latex]\,x\,[/latex]and[latex]\,y.\,[/latex]Solve the new equation for[latex]\,y.\,[/latex]The expression for[latex]\,y\,[/latex]is the inverse,[latex]\,y={f}^{-1}\left(x\right).[/latex]
Algebraic
Show that the function[latex]\,f\left(x\right)=a-x\,[/latex]is its own inverse for all real numbers[latex]\,a.\,[/latex]
For the following exercises, find[latex]\,{f}^{-1}\left(x\right)\,[/latex]for each function.
[latex]f\left(x\right)=x+3[/latex]
Show Solution
[latex]{f}^{-1}\left(x\right)=x-3[/latex]
[latex]f\left(x\right)=x+5[/latex]
[latex]f\left(x\right)=2-x[/latex]
Show Solution
[latex]{f}^{-1}\left(x\right)=2-x[/latex]
[latex]f\left(x\right)=3-x[/latex]
[latex]f\left(x\right)=\frac{x}{x+2}[/latex]
Show Solution
[latex]{f}^{-1}\left(x\right)=\frac{-2x}{x-1}[/latex]
[latex]f\left(x\right)=\frac{2x+3}{5x+4}[/latex]
For the following exercises, find a domain on which each function[latex]\,f\,[/latex]is one-to-one and non-decreasing. Write the domain in interval notation. Then find the inverse of[latex]\,f\,[/latex]restricted to that domain.
[latex]f\left(x\right)={\left(x+7\right)}^{2}[/latex]
Show Solution
domain of[latex]f\left(x\right):\,\left[-7,\infty \right);\,{f}^{-1}\left(x\right)=\sqrt{x}-7[/latex]
[latex]f\left(x\right)={\left(x-6\right)}^{2}[/latex]
[latex]f\left(x\right)={x}^{2}-5[/latex]
Show Solution
domain of[latex]\,f\left(x\right):\,\left[0,\infty \right);\,{f}^{-1}\left(x\right)=\sqrt{x+5}[/latex]
Given[latex]\,f\left(x\right)=\frac{x}{2+x}\,[/latex]and[latex]\,g\left(x\right)=\frac{2x}{1-x}:[/latex]
- Find[latex]\,f\left(g\left(x\right)\right)\,[/latex]and[latex]\,g\left(f\left(x\right)\right).[/latex]
- What does the answer tell us about the relationship between[latex]\,f\left(x\right)\,[/latex]and[latex]\,g\left(x\right)?[/latex]
Show Solution
a.[latex]\,f\left(g\left(x\right)\right)=x\,[/latex]and[latex]\,g\left(f\left(x\right)\right)=x.\,[/latex]b. This tells us that[latex]\,f\,[/latex]and[latex]\,g\,[/latex]are inverse functions
For the following exercises, use function composition to verify that[latex]\,f\left(x\right)\,[/latex]and[latex]\,g\left(x\right)\,[/latex]are inverse functions.
[latex]f\left(x\right)=\sqrt[3]{x-1}\,[/latex]and[latex]\,g\left(x\right)={x}^{3}+1[/latex]
Show Solution
[latex]f\left(g\left(x\right)\right)=x,\,g\left(f\left(x\right)\right)=x[/latex]
[latex]f\left(x\right)=-3x+5\,[/latex]and[latex]\,g\left(x\right)=\frac{x-5}{-3}[/latex]
Graphical
For the following exercises, use a graphing utility to determine whether each function is one-to-one.
Show Solution
one-to-one
[latex]f\left(x\right)=\sqrt[3]{3x+1}[/latex]
[latex]f\left(x\right)=-5x+1[/latex]
Show Solution
one-to-one
[latex]f\left(x\right)={x}^{3}-27[/latex]
For the following exercises, determine whether the graph represents a one-to-one function.
Show Solution
not one-to-one
For the following exercises, use the graph of[latex]\,f\,[/latex]shown in (Figure).
Find[latex]\,f\left(0\right).[/latex]
Show Solution
[latex]3[/latex]
Solve[latex]\,f\left(x\right)=0.[/latex]
Find[latex]\,{f}^{-1}\left(0\right).[/latex]
Show Solution
[latex]2[/latex]
Solve[latex]\,{f}^{-1}\left(x\right)=0.[/latex]
For the following exercises, use the graph of the one-to-one function shown in (Figure).
Sketch the graph of[latex]\,{f}^{-1}.\,[/latex]
Show Solution
Find[latex]\,f\left(6\right)\text{ and }{f}^{-1}\left(2\right).[/latex]
If the complete graph of[latex]\,f\,[/latex]is shown, find the domain of[latex]\,f.\,[/latex]
Show Solution
[latex]\left[2,10\right][/latex]
If the complete graph of[latex]\,f\,[/latex]is shown, find the range of[latex]\,f.[/latex]
Numeric
For the following exercises, evaluate or solve, assuming that the function[latex]\,f\,[/latex]is one-to-one.
If[latex]\,f\left(6\right)=7,\,[/latex]find[latex]\,\,{f}^{-1}\left(7\right).[/latex]
Show Solution
[latex]6[/latex]
If[latex]\,f\left(3\right)=2,\,[/latex]find[latex]\,{f}^{-1}\left(2\right).[/latex]
If[latex]\,{f}^{-1}\left(-4\right)=-8,\,[/latex]find[latex]\,f\left(-8\right).[/latex]
Show Solution
[latex]-4[/latex]
If[latex]\,{f}^{-1}\left(-2\right)=-1,\,[/latex]find[latex]\,f\left(-1\right).[/latex]
For the following exercises, use the values listed in (Figure) to evaluate or solve.
| [latex]x[/latex] | [latex]f\left(x\right)[/latex] |
|---|---|
| 0 | 8 |
| 1 | 0 |
| 2 | 7 |
| 3 | 4 |
| 4 | 2 |
| 5 | 6 |
| 6 | 5 |
| 7 | 3 |
| 8 | 9 |
| 9 | 1 |
Find[latex]\,f\left(1\right).[/latex]
Show Solution
[latex]0[/latex]
Solve[latex]\,f\left(x\right)=3.[/latex]
Find[latex]\,{f}^{-1}\left(0\right).[/latex]
Show Solution
[latex]\,1\,[/latex]
Solve[latex]\,{f}^{-1}\left(x\right)=7.[/latex]
Use the tabular representation of[latex]\,f\,[/latex]in (Figure) to create a table for[latex]\,{f}^{-1}\left(x\right).[/latex]
| [latex]x[/latex] | 3 | 6 | 9 | 13 | 14 |
| [latex]f\left(x\right)[/latex] | 1 | 4 | 7 | 12 | 16 |
Show Solution
| [latex]x[/latex] | 1 | 4 | 7 | 12 | 16 |
| [latex]{f}^{-1}\left(x\right)[/latex] | 3 | 6 | 9 | 13 | 14 |
Technology
For the following exercises, find the inverse function. Then, graph the function and its inverse.
[latex]f\left(x\right)=\frac{3}{x-2}[/latex]
[latex]f\left(x\right)={x}^{3}-1[/latex]
Show Solution
[latex]{f}^{-1}\left(x\right)={\left(1+x\right)}^{1/3}[/latex]
Find the inverse function of[latex]\,f\left(x\right)=\frac{1}{x-1}.\,[/latex]Use a graphing utility to find its domain and range. Write the domain and range in interval notation.
Real-World Applications
To convert from[latex]\,x\,[/latex]degrees Celsius to[latex]\,y\,[/latex]degrees Fahrenheit, we use the formula[latex]\,f\left(x\right)=\frac{9}{5}x+32.\,[/latex]Find the inverse function, if it exists, and explain its meaning.
Show Solution
[latex]{f}^{-1}\left(x\right)=\frac{5}{9}\left(x-32\right).\,[/latex]Given the Fahrenheit temperature,[latex]\,x,\,[/latex]this formula allows you to calculate the Celsius temperature.
The circumference[latex]\,C\,[/latex]of a circle is a function of its radius given by[latex]\,C\left(r\right)=2\pi r.\,[/latex]Express the radius of a circle as a function of its circumference. Call this function[latex]\,r\left(C\right).\,[/latex]Find[latex]\,r\left(36\pi \right)\,[/latex]and interpret its meaning.
A car travels at a constant speed of 50 miles per hour. The distance the car travels in miles is a function of time,[latex]\,t,\,[/latex]in hours given by[latex]\,d\left(t\right)=50t.\,[/latex]Find the inverse function by expressing the time of travel in terms of the distance traveled. Call this function[latex]\,t\left(d\right).\,[/latex]Find[latex]\,t\left(180\right)\,[/latex]and interpret its meaning.
Show Solution
[latex]t\left(d\right)=\frac{d}{50},\,[/latex][latex]t\left(180\right)=\frac{180}{50}.\,[/latex]The time for the car to travel 180 miles is 3.6 hours.
Chapter Review Exercises
Functions and Function Notation
For the following exercises, determine whether the relation is a function.
Show Solution
function
[latex]{y}^{2}+4=x,\,[/latex]for[latex]\,x\,[/latex]the independent variable and[latex]\,y\,[/latex]the dependent variable
Show Solution
not a function
Is the graph in (Figure) a function?
For the following exercises, evaluate the function at the indicated values:[latex]\,\,\,f\left(-3\right);\,\,f\left(2\right);\,\,\,f\left(-a\right);\,\,\,-f\left(a\right);\,\,\,f\left(a+h\right).[/latex]
[latex]f\left(x\right)=-2{x}^{2}+3x[/latex]
Show Solution
[latex]f\left(-3\right)=-27;[/latex][latex]f\left(2\right)=-2;[/latex][latex]f\left(-a\right)=-2{a}^{2}-3a;[/latex]
[latex]-f\left(a\right)=2{a}^{2}-3a;[/latex][latex]f\left(a+h\right)=-2{a}^{2}+3a-4ah+3h-2{h}^{2}[/latex]
[latex]f\left(x\right)=2|3x-1|[/latex]
For the following exercises, determine whether the functions are one-to-one.
[latex]f\left(x\right)=-3x+5[/latex]
Show Solution
one-to-one
[latex]f\left(x\right)=|x-3|[/latex]
For the following exercises, use the vertical line test to determine if the relation whose graph is provided is a function.
Show Solution
function
Show Solution
function
For the following exercises, graph the functions.
[latex]f\left(x\right)=|x+1|[/latex]
[latex]f\left(x\right)={x}^{2}-2[/latex]
Show Solution
For the following exercises, use (Figure) to approximate the values.
[latex]f\left(2\right)[/latex]
[latex]f\left(-2\right)[/latex]
Show Solution
[latex]2[/latex]
If[latex]\,f\left(x\right)=-2,\,[/latex]then solve for[latex]\,x.[/latex]
If[latex]\,f\left(x\right)=1,\,[/latex]then solve for[latex]\,x.[/latex]
Show Solution
[latex]x=-1.8\text{ }[/latex]or[latex]\text{ or }x=1.8[/latex]
For the following exercises, use the function[latex]\,h\left(t\right)=-16{t}^{2}+80t\,[/latex]to find the values in simplest form.
[latex]\frac{h\left(2\right)-h\left(1\right)}{2-1}[/latex]
[latex]\frac{h\left(a\right)-h\left(1\right)}{a-1}[/latex]
Show Solution
[latex]\frac{-64+80a-16{a}^{2}}{-1+a}=-16a+64[/latex]
Domain and Range
For the following exercises, find the domain of each function, expressing answers using interval notation.
[latex]f\left(x\right)=\frac{2}{3x+2}[/latex]
[latex]f\left(x\right)=\frac{x-3}{{x}^{2}-4x-12}[/latex]
Show Solution
[latex]\left(-\infty ,-2\right)\cup \left(-2,6\right)\cup \left(6,\infty \right)[/latex]
[latex]f\left(x\right)=\frac{\sqrt{x-6}}{\sqrt{x-4}}[/latex]
Graph this piecewise function:[latex]f\left(x\right)=\bigg\{\begin{array}{l}x+1\text{ }x<-2\\ -2x-3\text{ }x\ge -2\end{array}[/latex]
Show Solution
Rates of Change and Behavior of Graphs
For the following exercises, find the average rate of change of the functions from[latex]\,x=1\text{ to }x=2.[/latex]
[latex]f\left(x\right)=4x-3[/latex]
[latex]f\left(x\right)=10{x}^{2}+x[/latex]
Show Solution
[latex]31[/latex]
[latex]f\left(x\right)=-\frac{2}{{x}^{2}}[/latex]
For the following exercises, use the graphs to determine the intervals on which the functions are increasing, decreasing, or constant.
Show Solution
increasing[latex]\,\left(2,\infty \right);\,[/latex]
decreasing[latex]\,\left(-\infty ,2\right)[/latex]
Show Solution
increasing[latex]\text{}\left(-3,1\right);\text{}[/latex]constant[latex]\,\left(-\infty ,-3\right)\cup \left(1,\infty \right)[/latex]
Find the local minimum of the function graphed in (Figure).
Find the local extrema for the function graphed in (Figure).
Show Solution
local minimum[latex]\,\left(-2,-3\right);\,[/latex]local maximum[latex]\,\left(1,3\right)[/latex]
For the graph in (Figure), the domain of the function is[latex]\,\left[-3,3\right].[/latex]The range is[latex]\,\left[-10,10\right].\,[/latex]Find the absolute minimum of the function on this interval.
Find the absolute maximum of the function graphed in (Figure).
Show Solution
[latex]\,\left(-1.8,10\right)\,[/latex]
Composition of Functions
For the following exercises, find[latex]\,\left(f\circ g\right)\left(x\right)\,[/latex]and[latex]\,\left(g\circ f\right)\left(x\right)\,[/latex]for each pair of functions.
[latex]f\left(x\right)=4-x,\,g\left(x\right)=-4x[/latex]
[latex]f\left(x\right)=3x+2,\,g\left(x\right)=5-6x[/latex]
Show Solution
[latex]\left(f\circ g\right)\left(x\right)=17-18x;\,\left(g\circ f\right)\left(x\right)=-7-18x[/latex]
[latex]f\left(x\right)={x}^{2}+2x,\,g\left(x\right)=5x+1[/latex]
[latex]f\left(x\right)=\sqrt{x+2},\text{ }g\left(x\right)=\frac{1}{x}[/latex]
Show Solution
[latex]\left(f\circ g\right)\left(x\right)=\sqrt{\frac{1}{x}+2};\,\left(g\circ f\right)\left(x\right)=\frac{1}{\sqrt{x+2}}[/latex]
[latex]\,f\left(x\right)=\frac{x+3}{2},\text{ }g\left(x\right)=\sqrt{1-x}\,[/latex]
For the following exercises, find[latex]\,\left(f\circ g\right)\,[/latex]and the domain for[latex]\,\left(f\circ g\right)\left(x\right)\,[/latex]for each pair of functions.
[latex]f\left(x\right)=\frac{x+1}{x+4},\text{ }g\left(x\right)=\frac{1}{x}[/latex]
Show Solution
[latex]\left(f\circ g\right)\left(x\right)=\frac{1+x}{1+4x}, x\ne 0, x\ne -\frac{1}{4}[/latex]
[latex]f\left(x\right)=\frac{1}{x+3},\text{ }g\left(x\right)=\frac{1}{x-9}[/latex]
[latex]f\left(x\right)=\frac{1}{x},\text{ }g\left(x\right)=\sqrt{x}[/latex]
Show Solution
[latex]\left(f\circ g\right)\left(x\right)=\frac{1}{\sqrt{x}},\,x>0[/latex]
[latex]f\left(x\right)=\frac{1}{{x}^{2}-1},\text{ }g\left(x\right)=\sqrt{x+1}[/latex]
For the following exercises, express each function[latex]\,H\,[/latex]as a composition of two functions[latex]\,f\,[/latex]and[latex]\,g\,[/latex]where[latex]\,H\left(x\right)=\left(f\circ g\right)\left(x\right).[/latex]
[latex]H\left(x\right)=\sqrt{\frac{2x-1}{3x+4}}[/latex]
Show Solution
sample:[latex]\,g\left(x\right)=\frac{2x-1}{3x+4};\,f\left(x\right)=\sqrt{x}[/latex]
[latex]H\left(x\right)=\frac{1}{{\left(3{x}^{2}-4\right)}^{-3}}[/latex]
Transformation of Functions
For the following exercises, sketch a graph of the given function.
[latex]f\left(x\right)={\left(x-3\right)}^{2}[/latex]
Show Solution
[latex]f\left(x\right)={\left(x+4\right)}^{3}[/latex]
[latex]f\left(x\right)=\sqrt{x}+5[/latex]
Show Solution
[latex]f\left(x\right)=-{x}^{3}[/latex]
[latex]f\left(x\right)=\sqrt[3]{-x}[/latex]
Show Solution
[latex]f\left(x\right)=5\sqrt{-x}-4[/latex]
[latex]f\left(x\right)=4\left[|x-2|-6\right][/latex]
Show Solution
[latex]f\left(x\right)=-{\left(x+2\right)}^{2}-1[/latex]
For the following exercises, sketch the graph of the function[latex]\,g\,[/latex]if the graph of the function[latex]\,f\,[/latex]is shown in (Figure).
[latex]g\left(x\right)=f\left(x-1\right)[/latex]
Show Solution
[latex]g\left(x\right)=3f\left(x\right)[/latex]
For the following exercises, write the equation for the standard function represented by each of the graphs below.
Show Solution
[latex]f\left(x\right)=|x-3|[/latex]
For the following exercises, determine whether each function below is even, odd, or neither.
[latex]f\left(x\right)=3{x}^{4}[/latex]
Show Solution
even
[latex]g\left(x\right)=\sqrt{x}[/latex]
[latex]h\left(x\right)=\frac{1}{x}+3x[/latex]
Show Solution
odd
For the following exercises, analyze the graph and determine whether the graphed function is even, odd, or neither.
Show Solution
even
Absolute Value Functions
For the following exercises, write an equation for the transformation of[latex]\,f\left(x\right)=|x|.[/latex]
Show Solution
[latex]f\left(x\right)=\frac{1}{2}|x+2|+1[/latex]
Show Solution
[latex]f\left(x\right)=-3|x-3|+3[/latex]
For the following exercises, graph the absolute value function.
[latex]f\left(x\right)=|x-5|[/latex]
[latex]f\left(x\right)=-|x-3|[/latex]
Show Solution
[latex]f\left(x\right)=|2x-4|[/latex]
Inverse Functions
For the following exercises, find[latex]\text{ }{f}^{-1}\left(x\right)\text{ }[/latex]for each function.
[latex]f\left(x\right)=9+10x[/latex]
[latex]f\left(x\right)=\frac{x}{x+2}[/latex]
Show Solution
[latex]{f}^{-1}\left(x\right)=\frac{-2x}{x-1}[/latex]
For the following exercise, find a domain on which the function[latex]\text{ }f\text{ }[/latex]is one-to-one and non-decreasing. Write the domain in interval notation. Then find the inverse of[latex]\text{ }f\text{ }[/latex]restricted to that domain.
[latex]f\left(x\right)={x}^{2}+1[/latex]
Given [latex]f\left(x\right)={x}^{3}-5[/latex] and [latex]g\left(x\right)=\sqrt[3]{x+5}:[/latex]
- Find [latex]f\left(g\left(x\right)\right)[/latex] and [latex]g\left(f\left(x\right)\right).[/latex]
- What does the answer tell us about the relationship between [latex]f\left(x\right)[/latex] and [latex]g\left(x\right)?[/latex]
Show Solution
- [latex]f\left(g\left(x\right)\right)=x[/latex] and [latex]g\left(f\left(x\right)\right)=x.[/latex]
- This tells us that [latex]f[/latex] and [latex]g[/latex] are inverse functions
For the following exercises, use a graphing utility to determine whether each function is one-to-one.
Show Solution
The function is one-to-one.
[latex]f\left(x\right)=-3{x}^{2}+x[/latex]
Show Solution
The function is not one-to-one.
If [latex]f\left(5\right)=2,[/latex] find [latex]{f}^{-1}\left(2\right).[/latex]
Show Solution
[latex]5[/latex]
If [latex]f\left(1\right)=4,[/latex] find [latex]{f}^{-1}\left(4\right).[/latex]
Practice Test
For the following exercises, determine whether each of the following relations is a function.
[latex]y=2x+8[/latex]
Show Solution
The relation is a function.
[latex]\left\{\left(2,1\right),\left(3,2\right),\left(-1,1\right),\left(0,-2\right)\right\}[/latex]
For the following exercises, evaluate the function[latex]\,f\left(x\right)=-3{x}^{2}+2x\,[/latex]
at the given input.
Show Solution
−16
[latex]\,f\left(a\right)\,[/latex]
Show that the function[latex]\,f\left(x\right)=-2{\left(x-1\right)}^{2}+3\,[/latex]is not one-to-one.
Show Solution
The graph is a parabola and the graph fails the horizontal line test.
Write the domain of the function[latex]\,f\left(x\right)=\sqrt{3-x}\,[/latex]in interval notation.
Given[latex]\,f\left(x\right)=2{x}^{2}-5x,\,[/latex]find[latex]f\left(a+1\right)-f\left(1\right)\,[/latex]in simplest form.
Show Solution
[latex]2{a}^{2}-a[/latex]
Graph the function[latex]f\left(x\right)=\left\{\begin{array}{cc}x+1\text{ if}& -2
Find the average rate of change of the function[latex]\,f\left(x\right)=3-2{x}^{2}+x\,[/latex]by finding[latex]\,\frac{f\left(b\right)-f\left(a\right)}{b-a}\,[/latex]in simplest form.
Show Solution
[latex]-2\left(a+b\right)+1[/latex]
For the following exercises, use the functions[latex]\,f\left(x\right)=3-2{x}^{2}+x\text{ and }g\left(x\right)=\sqrt{x}\,[/latex]to find the composite functions.
[latex]\left(g\circ f\right)\left(x\right)[/latex]
[latex]\left(g\circ f\right)\left(1\right)[/latex]
Show Solution
[latex]\sqrt{2}[/latex]
Express[latex]\,H\left(x\right)=\sqrt[3]{5{x}^{2}-3x}\,[/latex]as a composition of two functions,[latex]\,f\,[/latex]and[latex]\,g,\,[/latex]where[latex]\,\left(f\circ g\right)\left(x\right)=H\left(x\right).[/latex]
For the following exercises, graph the functions by translating, stretching, and/or compressing a toolkit function.
[latex]f\left(x\right)=\sqrt{x+6}-1[/latex]
Show Solution
[latex]f\left(x\right)=\frac{1}{x+2}-1[/latex]
For the following exercises, determine whether the functions are even, odd, or neither.
[latex]f\left(x\right)=-\frac{5}{{x}^{2}}+9{x}^{6}[/latex]
Show Solution
[latex]\text{even}[/latex]
[latex]f\left(x\right)=\frac{1}{x}[/latex]
Show Solution
[latex]\text{odd}[/latex]
Graph the absolute value function[latex]\,f\left(x\right)=-2|x-1|+3.[/latex]
For the following exercises, find the inverse of the function.
[latex]f\left(x\right)=3x-5[/latex]
Show Solution
[latex]{f}^{-1}\left(x\right)=\frac{x+5}{3}[/latex]
[latex]f\left(x\right)=\frac{4}{x+7}[/latex]
For the following exercises, use the graph of[latex]\,g\,[/latex]shown in (Figure).
On what intervals is the function increasing?
Show Solution
[latex]\left(-\infty ,-1.1\right)\text{ and }\left(1.1,\infty \right)[/latex]
On what intervals is the function decreasing?
Approximate the local minimum of the function. Express the answer as an ordered pair.
Show Solution
[latex]\left(1.1,-0.9\right)[/latex]
Approximate the local maximum of the function. Express the answer as an ordered pair.
For the following exercises, use the graph of the piecewise function shown in (Figure).
Find[latex]\,f\left(2\right).[/latex]
Show Solution
[latex]f\left(2\right)=2[/latex]
Find[latex]\,f\left(-2\right).[/latex]
Write an equation for the piecewise function.
Show Solution
[latex]f\left(x\right)=\left\{\begin{array}{c}|x|\,\,\,\text{if}\,\,x\le 2\\ 3\,\,\,\,\,\text{if}\,\,x>2\end{array}[/latex]
For the following exercises, use the values listed in (Figure).
| [latex]x[/latex] | [latex]F\left(x\right)[/latex] |
|---|---|
| 0 | 1 |
| 1 | 3 |
| 2 | 5 |
| 3 | 7 |
| 4 | 9 |
| 5 | 11 |
| 6 | 13 |
| 7 | 15 |
| 8 | 17 |
Find[latex]\,F\left(6\right).[/latex]
Solve the equation[latex]\,F\left(x\right)=5.[/latex]
Show Solution
[latex]x=2[/latex]
Is the graph increasing or decreasing on its domain?
Is the function represented by the graph one-to-one?
Show Solution
yes
Find[latex]\,{F}^{-1}\left(15\right).[/latex]
Given[latex]\,f\left(x\right)=-2x+11,\,[/latex]find[latex]\,{f}^{-1}\left(x\right).[/latex]
Show Solution
[latex]{f}^{-1}\left(x\right)=-\frac{x-11}{2}[/latex]
Glossary
- inverse function
- for any one-to-one function[latex]\,f\left(x\right),\,[/latex]the inverse is a function[latex]\,{f}^{-1}\left(x\right)\,[/latex]such that[latex]\,{f}^{-1}\left(f\left(x\right)\right)=x\,[/latex]for all[latex]\,x\,[/latex]in the domain of[latex]\,f;\,[/latex]this also implies that[latex]\,f\left({f}^{-1}\left(x\right)\right)=x\,[/latex]for all[latex]\,x\,[/latex]in the domain of[latex]\,{f}^{-1}[/latex]
