Problem 12
Question
The functions in Exercises \(11-30\) are all one-to-one. For each function: a. Find an equation for \(f^{-1}(x),\) the inverse function. b. Verify that your equation is correct by showing that $$f\left(f^{-1}(x)\right)=x \text { and } f^{-1}(f(x))=x$$ $$f(x)=x+5$$
Step-by-Step Solution
Verified Answer
The inverse function of \(f(x) = x + 5\) is \(f^{-1}(x) = x - 5\), and it is verified by showing that \(f(f^{-1}(x)) = x\) and \(f^{-1}(f(x)) = x\).
1Step 1: Find the Inverse Function
To find the inverse of the function \(f(x) = x + 5\), replace \(f(x)\) with \(y\). So we have \(y = x + 5\). Now, switch \(x\) and \(y\) and solve for \(y\), giving us \(x = y + 5\) or \(y = x - 5\). This means the inverse function \(f^{-1}(x) = x - 5.\)
2Step 2: Verify the Inverse Function (part a)
But, we need to make sure that our calculated inverse is correct. Substitute \(f^{-1}(x)\) into \(f(x)\). We get \(f(f^{-1}(x)) = f(x - 5) = (x - 5) + 5 = x\). This verifies the first part as \(f(f^{-1}(x)) = x\).
3Step 3: Verify the Inverse Function (part b)
Now, substitute \(f(x)\) into \(f^{-1}(x)\). We get \(f^{-1}(f(x)) = f^{-1}(x + 5) = (x + 5) - 5 = x\). This verifies the second part as \(f^{-1}(f(x)) = x\). Thus, our calculated inverse function is correct.
Key Concepts
Understanding One-to-One FunctionsFunction Verification TechniquesAlgebraic Manipulation Steps for Inverses
Understanding One-to-One Functions
A one-to-one function is a special type of function where each input has a unique output. This means no two different inputs in the domain are mapped to the same output in the range. One-to-one functions are also known as injective functions. They play a crucial role when it comes to finding inverse functions because only one-to-one functions have inverses that themselves are functions.
Some main characteristics of one-to-one functions include:
- Every element in the domain is mapped to a unique element in the range.
- The horizontal line test can be used to determine if a function is one-to-one. If any horizontal line intersects the graph of the function at most once, then the function is one-to-one.
Function Verification Techniques
Verifying a function, particularly when dealing with inverses, ensures the calculations accurately represent the inverse relationship. In the provided exercise, two verification techniques are used to check the authenticity of the inverse function. To verify that your inverse is correct, you should:
- Substitute the inverse function back into the original function and check if the result is the input variable, like in the equation: \(f(f^{-1}(x)) = x\).
- Substitute the original function into the inverse function and similarly check if the result is the input variable, which is demonstrated by: \(f^{-1}(f(x)) = x\).
Algebraic Manipulation Steps for Inverses
Algebraic manipulation involves rearranging equations to solve for a particular variable. This skill is especially helpful when finding inverse functions. Here’s a breakdown of the algebraic steps involved:
- Start by replacing the function notation \(f(x)\) with \(y\). For example, \(y = x + 5\).
- Swap the variables, effectively exchanging \(x\) and \(y\) to prepare to solve for the inverse. This gives you \(x = y + 5\).
- Solve the equation for \(y\) to express it in terms of \(x\). So for our example, solving \(x = y + 5\) gives us \(y = x - 5\).
- Recognize \(y\) as \(f^{-1}(x)\) so the inverse function is \(f^{-1}(x) = x - 5\).
Other exercises in this chapter
Problem 12
Use the given conditions to write an equation for each line in point-slope form and slope-intercept form. Slope \(=4,\) passing through \((1,3)\)
View solution Problem 12
Find \(f+g, f-g, f g,\) and \(\frac{f}{g}\). Determine the domain for each function. $$f(x)=\frac{1}{x}, g(x)=x-5$$
View solution Problem 12
In Exercises \(9-20,\) determine whether each equation defines \(y\) as a function of \(x .\) $$x^{2}+y=25$$
View solution Problem 12
Find the distance between each pair of points. If necessary, round answers to two decimals places. $$ (0,-\sqrt{3}) \text { and }(\sqrt{5}, 0) $$
View solution