Problem 30
Question
Inverse Functions In Exercises \(29-32\) , illustrate that the functions are inverses of each other by graphing both functions on the same set of coordinate axes. $$ \begin{array}{l}{f(x)=e^{x / 3}} \\ {g(x)=\ln x^{3}}\end{array} $$
Step-by-Step Solution
Verified Answer
From the visualized graphs, it can be concluded that the functions \(f(x)=e^{x/3}\) and \(g(x)=\ln{x^{3}}\) are inverses of each other, because the graph of \(g(x)\) is the reflection of the graph of \(f(x)\) about the line \(y=x\). This shows that they 'undo' each other, verifying the property of inverse functions.
1Step 1: Identify the inverse of each function
In general, the inverse function of \(f(x)\) is denoted as \(f^{-1}(x)\). This means that for \(f(x)=e^{x/3}\), the inverse is \(f^{-1}(x)=\ln{x}^{3}\). Similarly, for \(g(x)=\ln{x^{3}}\), the inverse is \(g^{-1}(x)=e^{x/3}\). It can be seen that the inverse of \(f\) is \(g\) and vice versa, validating the inverse function relationship.
2Step 2: Plot the graphs of each function
To visualize the inverse relationship, plot both \(f(x)\) and \(g(x)\) on the same coordinate axis. Because \(f(x)=e^{x/3}\) and \(g(x)=\ln{x^{3}}\) are exponential and logarithmic functions respectively, they both exhibit growth, but at different rates. The graph of \(f(x)\) starts at \(y=1\) for \(x=0\), then increases as \(x\) increases. Similarly, the graph of \(g(x)\) starts at \(y=0\) for \(x=1\), then increases as \(x\) increases.
3Step 3: Illustrate the inverse property from the graph
On the graph, the functions \(f(x)\) and \(g(x)\) should be mirror images of one another about the line \(y=x\). This is because any point \((x,y)\) on the graph of \(f(x)\) has an equivalent point \((y,x)\) on the graph of \(g(x)\), and vice versa, which shows the property of inverse functions -- they 'undo' each other.
Key Concepts
Understanding Exponential FunctionsExploring Logarithmic FunctionsGraph Transformations and Inverse Relationships
Understanding Exponential Functions
Exponential functions are a fundamental concept in mathematics, often used to describe growth processes, whether in populations, investments, or even radioactive decay. An exponential function can be generally expressed in the form \(f(x) = a \, e^{bx}\), where \(e\) is approximately 2.718, known as Euler's number. This function translates constant differences in the independent variable, \(x\), into constant ratios in the dependent variable, \(f(x)\).
In the exercise, we explored the function \(f(x) = e^{x/3}\). Here, \(x\) is divided by 3 before being plugged into the exponential function, which has the effect of stretching the graph horizontally, making the curve shallower than the standard exponential function \(y = e^x\).
Recognizing how changes in this formula affect the curve is crucial. When \(a = 1\) and \(b = \frac{1}{3}\), as in our exercise, the initial value for \(f(x)\) at \(x = 0\) is 1, and the function increases as \(x\) increases, representing an exponential growth function with a shallower curve.
In the exercise, we explored the function \(f(x) = e^{x/3}\). Here, \(x\) is divided by 3 before being plugged into the exponential function, which has the effect of stretching the graph horizontally, making the curve shallower than the standard exponential function \(y = e^x\).
Recognizing how changes in this formula affect the curve is crucial. When \(a = 1\) and \(b = \frac{1}{3}\), as in our exercise, the initial value for \(f(x)\) at \(x = 0\) is 1, and the function increases as \(x\) increases, representing an exponential growth function with a shallower curve.
Exploring Logarithmic Functions
Logarithmic functions are the inverses of exponential functions and play a significant role when dealing with inverse relationships, such as in our exercise. The base-\(e\) logarithm, which is the natural logarithm, is denoted as \(\ln\). So, a logarithmic function can be expressed as \(g(x) = \ln{(x^3)}\).
In the exercise, this function is the inverse of the exponential function \(f(x) = e^{x/3}\). It is important to understand that logarithms convert multiplicative relationships into additive ones, simplifying many mathematical processes. The power of a positive base raised to a variable is computed by its logarithm, which matches the variable.
Graphically, the function \(g(x) = \ln{x}^3\) begins at \(y = 0\) when \(x = 1\), then rises as \(x\) increases. This upward sloping trend represents growth; however, it does so in a more moderated way compared to its exponential counterpart. Understanding how logarithms help in reversing exponential trends is crucial when interpreting data that initially appears non-linear.
In the exercise, this function is the inverse of the exponential function \(f(x) = e^{x/3}\). It is important to understand that logarithms convert multiplicative relationships into additive ones, simplifying many mathematical processes. The power of a positive base raised to a variable is computed by its logarithm, which matches the variable.
Graphically, the function \(g(x) = \ln{x}^3\) begins at \(y = 0\) when \(x = 1\), then rises as \(x\) increases. This upward sloping trend represents growth; however, it does so in a more moderated way compared to its exponential counterpart. Understanding how logarithms help in reversing exponential trends is crucial when interpreting data that initially appears non-linear.
Graph Transformations and Inverse Relationships
Graph transformations are valuable tools that help visualize changes in functions and their inverses. For instance, transforming the graph of \(f(x) = e^{x/3}\) into its inverse \(g(x) = \ln{x^3}\) is about flipping the graph over the line \(y = x\). This reflects each point \((x,y)\) on the graph of \(f(x)\) to \((y,x)\) on the graph of \(g(x)\).
To better understand this, remember:
To better understand this, remember:
- Any point \((a,b)\) on a function's graph becomes \((b,a)\) on the graph of its inverse.
- The line \(y = x\) acts as a mirror, flipping the graphs accordingly.
- This mirroring shows that each function "undoes" the actions of the other.
Other exercises in this chapter
Problem 30
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