Problem 26 - College Algebra

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 $$f\left(f^{-1}(x)\right)=x \text { and } f^{-1}(f(x))=x$$ $$f(x)=x^{2}-1, \text { for } x \geq 0$$

Step-by-Step Solution

Answer

Thus, the inverse function \(f^{-1}(x)\) of given function \(f(x)=x^{2}-1\) for \(x \geq 0\) is \(f^{-1}(x)=\sqrt{x+1}\), and it has been verified using the properties \(f(f^{-1}(x))=x\) and \(f^{-1}(f(x))=x\).

Step 1: Find the Inverse Function

To find the inverse of the function \(f(x)=x^{2}-1\) for \(x \geq 0\), switch 'x' and 'y', so we have \(x=y^{2}-1\). Solve this for 'y' to find the equation of the inverse function. Start by adding 1 on both sides, giving \(x+1=y^{2}\), then square root both sides, leading to \(y=\sqrt{x+1}\). Because we are working with the nonnegative scope of the function, it is the positive square root. So, the inverse function, denoted \(f^{-1}(x)\), is \(f^{-1}(x)=\sqrt{x+1}\).

Step 2: Verifying the Inverse Function

To prove that \(f^{-1}(x)\) is indeed the inverse of \(f(x)\), show that \(f(f^{-1}(x))=x\) and \(f^{-1}(f(x))=x\).For this, start by substituting \(f^{-1}(x)\) into the function \(f(x)\): \(f(f^{-1}(x))=f(\sqrt{x+1})=(\sqrt{x+1})^{2}-1=x+1-1=x\). So, \(f(f^{-1}(x))=x\) holds.Next, substitute \(f(x)\) into \(f^{-1}(x)\): \(f^{-1}(f(x))=f^{-1}(x^{2}-1)=\sqrt{(x^{2}-1)+1}=\sqrt{x^{2}}=x\), since for all \(x \geq 0\), \(x=\sqrt{x^{2}}\). Thus, \(f^{-1}(f(x))=x\) also holds.

Key Concepts

one-to-one functions

A function is described as one-to-one if every y-value in the function's range is paired with exactly one unique x-value in its domain. In simpler terms, no two different x-values produce the same y-value.
This characteristic is essential when determining whether a function has an inverse. The inverse of a function essentially "reverses" the roles of the x and y values, flipping the function over the line y = x. Without being one-to-one, a function cannot pass this reversal uniquely.

• How to Check for a One-to-One Function: Use the Horizontal Line Test by drawing horizontal lines across the graph of the function. If any horizontal line crosses the graph more than once, the function isn't one-to-one.
• The function \(f(x) = x^2 - 1\) for \(x \geq 0\) is a good example. Here, this restriction makes sure it satisfies being one-to-one for nonnegative x values.

verifying inverse functions

Once you've identified an inverse function, it's crucial to verify it using two main conditions: \(f(f^{-1}(x)) = x\) and \(f^{-1}(f(x)) = x\).
This process ensures that the inverse truly reverses the original function without losing or mixing up any values.

• First Verification: This involves plugging the inverse into the original function and simplifying. If you return to your starting x-value, the condition \(f(f^{-1}(x)) = x\) is satisfied.
• Second Verification: Swap the inputs by placing the original function into the inverse. Simplifying back to your original x shows that \(f^{-1}(f(x)) = x\) also holds true.
• For example, in the exercise, by using \(f(x) = x^2 - 1\) and its inverse \(f^{-1}(x) = \sqrt{x+1}\), both verifications prove the functions undo each other effectively.

finding inverse equations

Finding the inverse of a function involves reversing the process of the original function equation. This swaps the roles of x and y and solves for the new y.
To do this, follow these steps:

• Switch x and y: Begin with your function typically written as \(y = f(x)\). Switch it to \(x = f(y)\).
• Solve for y: After switching, solve the equation for y. This gives you the inverse function \(f^{-1}(x)\).
• Consider context and restrictions. For example, in \(f(x) = x^2 - 1\), restraining the domain ensures an inverse since only nonnegative x-values are used. This leads to \(f^{-1}(x) = \sqrt{x+1}\).

Remember, not all functions will have inverses that are simple to find or express cleanly, so attention to the domain and function type is crucial.