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Magazine Chapter 12: Problem 47
Solve. If \(f(x)=2 x+1,\) show that \(f^{-1}(x)=\frac{x-1}{2}\).
Short Answer
Expert verified
The inverse is \(f^{-1}(x) = \frac{x-1}{2}\), and it is correct.
Step by step solution
01
Understanding the Function and Its Inverse
The function given is \(f(x) = 2x + 1\). We need to find the inverse, \(f^{-1}(x)\), such that when \(f\) and \(f^{-1}\) are composed, they return the identity function, i.e., \(f(f^{-1}(x)) = x\) and \(f^{-1}(f(x)) = x\).
02
Set Up the Equation for the Inverse
To find \(f^{-1}(x)\), we swap \(f(x)\) and \(x\). So, start by setting \(y = f(x) = 2x + 1\). Now solve for \(x\) in terms of \(y\) by the equation \(y = 2x + 1\).
03
Solve for x in Terms of y
Rearrange the equation \(y = 2x + 1\) to solve for \(x\). First subtract 1 from both sides, giving \(y - 1 = 2x\). Then divide both sides by 2: \(x = \frac{y - 1}{2}\).
04
Write the Inverse Function
Now that we have \(x\) in terms of \(y\), we can state the inverse function. Replace \(y\) with \(x\) in the equation \(x = \frac{y - 1}{2}\) to find \(f^{-1}(x) = \frac{x - 1}{2}\).
05
Verify the Inverse
To confirm that \(f^{-1}(x) = \frac{x - 1}{2}\) is indeed correct, compose it with \(f(x)\): \(f(f^{-1}(x)) = 2\left(\frac{x - 1}{2}\right) + 1 = x - 1 + 1 = x\).Compose the inverse with the original function as well: \(f^{-1}(f(x)) = \frac{2x + 1 - 1}{2} = x\).Both compositions result in \(x\), confirming correct inverses.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Function Composition
Function composition is like a well-choreographed dance where two functions are combined to form a new one. Suppose we have two functions, \( f(x) \) and \( g(x) \). When we compose them, we create a new function \( h(x) \) defined as \( h(x) = f(g(x)) \). Here, \( g(x) \) outputs a result that immediately becomes the input for \( f(x) \).
- First, compute the inner function \( g(x) \).
- Next, plug the result of \( g(x) \) into the outer function \( f(x) \).
Solving Equations
Solving equations is like solving a puzzle. You start with an equation and your task is to find the value of the variable that makes the equation true. For this exercise, we're dealing with a linear equation.
To find an inverse, you need to resolve the original function equation by switching the variables. For example:
To find an inverse, you need to resolve the original function equation by switching the variables. For example:
- Given \( y = 2x + 1 \), you aim to solve for \( x \).
- Reorganize the equation, so it reads \( x = \frac{y-1}{2} \).
Identity Function
Imagine the identity function as a universal truth in algebra, which simply states \( f(x) = x \). This function doesn鈥檛 change its input and returns it as is. The significance of the identity function lies in its role within the composition of a function and its inverse.
When you verify that two functions鈥攕ay \( f(x) \) and its inverse \( f^{-1}(x) \)鈥攁re genuinely inverses, you do so by composing them and getting back the identity function:
When you verify that two functions鈥攕ay \( f(x) \) and its inverse \( f^{-1}(x) \)鈥攁re genuinely inverses, you do so by composing them and getting back the identity function:
- \( f(f^{-1}(x)) = x \),
- \( f^{-1}(f(x)) = x \).
Algebraic Manipulation
Algebraic manipulation is a magician's performance in mathematics where you rearrange and adjust an equation's structure to solve it or rewrite it. In our exercise, finding the inverse of \( f(x) = 2x + 1 \) required a series of intuitive algebraic maneuvers:
- Start with \( y = 2x + 1 \).
- Subtract 1: \( y - 1 = 2x \).
- Divide by 2: \( x = \frac{y - 1}{2} \).
