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Magazine Chapter 14: Problem 9
Show that \(g(x)=\pi \log _{2}(\pi x)-\pi^{2}\) is invertible. Find \(g^{-1}(x)\).
Short Answer
Expert verified
Yes, the function \(g(x)=\pi \log _{2}(\pi x)-\pi^{2}\) is invertible. And its inverse function is \(g^{-1}(x) = 2^{(x + \pi^{2})/\pi}/\pi\).
Step by step solution
01
Prove the function is one-to-one
A function is one-to-one if every x-value has a unique y-value. In this case, take two random values \(x_1\) and \(x_2\). If \(g(x_1) = g(x_2)\), then it should imply \(x_1 = x_2\). Now we can see that if \(\pi \log _{2}(\pi x_1)-\pi^{2} = \pi \log _{2}(\pi x_2)-\pi^{2}\), it can be simplified to \(\log _{2}(\pi x_1) = \log _{2}(\pi x_2)\). Using laws of logarithms, this equation further simplifies to \(x_1 = x_2\). Hence, the function is one-to-one.
02
Prove the function is onto
A function is onto if every y-value is mapped by some x-value. This is generally more difficult to prove. But look at the given function, we have a logarithm function which is defined for all the real numbers and -pi squared which is a constant. Hence the function is defined for all real numbers. This implies the function is onto.
03
Find the Inverse
As the function is both one-to-one and onto, it's clear that the function is invertible. So, to find \(g^{-1}(x)\), first replace \(g(x)\) with y, which gives \(y = \pi \log _{2}(\pi x)-\pi^{2}\). Solve this equation for x, first move \(\pi^{2}\) to the left side, then divide both sides by \(\pi\) to get \(\log _{2}(\pi x) = (y + \pi^{2})/\pi\), then convert to exponential form that gives \(\pi x = 2^{(y + \pi^{2})/\pi}\), finally solve for x to get \(x = 2^{(y + \pi^{2})/\pi}/\pi\), replace y with \(x\) in \(\pi x\) to get the inverse function \(g^{-1}(x) = 2^{(x + \pi^{2})/\pi}/\pi\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
One-to-One Function
Understanding what a one-to-one function entails is essential when examining whether a function has an inverse. This characteristic means that each element in the domain (input) of the function corresponds to a unique element in the codomain (output). Imagine a scenario where each person is assigned a unique ID number; similarly, in a one-to-one function, each input has a distinct output.
The textbook solution demonstrated that for the function given, no two different inputs produce the same output, which echoes the real-world example of unique IDs. This exclusivity is crucial for invertibility because it ensures that the inverse function can map each output back to one specific input without any ambiguity.
The textbook solution demonstrated that for the function given, no two different inputs produce the same output, which echoes the real-world example of unique IDs. This exclusivity is crucial for invertibility because it ensures that the inverse function can map each output back to one specific input without any ambiguity.
Onto Function
Conversely, an onto function, also known as a surjective function, ensures that every possible output is connected to at least one input. Think of an onto function as a busy receptionist making sure every call is answered - no call (output) is left unattended.
By proving the function is onto, we make sure that the domain covers the entire range of possible outputs. The example used in the textbook solution aligns with this concept, as it indicates that the logarithmic part of the function can take any real number, ensuring the function 'reaches out' to cover all possible values within its range.
By proving the function is onto, we make sure that the domain covers the entire range of possible outputs. The example used in the textbook solution aligns with this concept, as it indicates that the logarithmic part of the function can take any real number, ensuring the function 'reaches out' to cover all possible values within its range.
Logarithmic Functions
Logarithmic functions are the inverses of exponential functions and play a significant role in mathematics. They can be identified by their signature format, \( y = \log_b(x) \), where \b\ is the base of the logarithm. These functions enable us to untangle exponential growth or decay problems, making them particularly useful in fields such as finance and natural sciences.
A key property of logarithms that's useful for solving equations is that they can convert multiplications into additions and divisions into subtractions. This helps in simplifying complex exponential relations, like in our exercise where we manipulate logarithms to solve for the inverse function.
A key property of logarithms that's useful for solving equations is that they can convert multiplications into additions and divisions into subtractions. This helps in simplifying complex exponential relations, like in our exercise where we manipulate logarithms to solve for the inverse function.
Inverse Functions
Thinking about inverse functions can be similar to considering a two-way street. Just as you can travel from A to B and then back again, an inverse function allows you to start with the output of a function and find the original input. If a function represents how to get from your home to your friend's house, the inverse function would be the route from your friend's house back to yours.
The steps to find the inverse function often involve swapping the 'x' and 'y' in the original function and then solving for 'y'. In our example with the function \(g(x)\), after proving it is both one-to-one and onto, we follow these steps to successfully find its inverse \(g^{-1}(x)\).
The steps to find the inverse function often involve swapping the 'x' and 'y' in the original function and then solving for 'y'. In our example with the function \(g(x)\), after proving it is both one-to-one and onto, we follow these steps to successfully find its inverse \(g^{-1}(x)\).
Properties of Logarithms
Logarithms have unique properties that are tools to simplify expressions and solve equations involving exponential terms. For instance, consider the following key properties:
Understanding and applying these properties make it possible to maneuver between logarithmic and exponential forms seamlessly, which is exactly what was required to derive the inverse function in the provided exercise.
- The Product Rule: \( \log_b(m \cdot n) = \log_b(m) + \log_b(n) \), depicting how to break down a multiplication into an addition.
- The Quotient Rule: \( \log_b(\frac{m}{n}) = \log_b(m) - \log_b(n) \), which showcases the transformation of a division into a subtraction.
- The Power Rule: \( \log_b(m^n) = n \cdot \log_b(m) \), provides a way to handle an exponent by bringing it out in front of the logarithm.
Understanding and applying these properties make it possible to maneuver between logarithmic and exponential forms seamlessly, which is exactly what was required to derive the inverse function in the provided exercise.
