Problem 25 - 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

The inverse function of \(f(x)=x^{2}+1, \text { for } x \geq 0\) is \(f^{-1}(x)=\sqrt{x-1}\).

Step 1: Finding the Inverse Function

To find the inverse of a function, we start by replacing \(f(x)\) by \(y\). So, we have the equation \(y=x^{2}+1\). Then, we swap \(x\) and \(y\) to get \(x=y^{2}+1\). Solving this equation for \(y\) will give us the inverse function. Subtract 1 from both sides to get \(x-1=y^{2}\). As \(y\) is positive given \(x \geq 0\), we take the positive square root on both sides to get \(y=\sqrt{x-1}\). This gives us the inverse function \(f^{-1}(x)=\sqrt{x-1}\).

Step 2: Verifying the Inverse Function

To verify the inverse function, we need to prove that \(f(f^{-1}(x))=x\) and \(f^{-1}(f(x)) = x\). Substituting \(f^{-1}(x)=\sqrt{x-1}\) into \(f(x)\) gives us \(f(f^{-1}(x))=(\sqrt{x-1})^2+1\). Simplifying this, we get \(f(f^{-1}(x))=x-1+1=x\). Substituting \(f(x)=x^2+1\) into \(f^{-1}(x)\) gives us \(f^{-1}(f(x))=\sqrt{(x^2+1)-1}\). Simplifying this, we get \(f^{-1}(f(x))=\sqrt{x^2}=x\), which holds true for \(x \geq 0\). Hence, our inverse function is indeed the correct one.

Step 3: Conclusion

The inverse of the function \(f(x)=x^{2}+1, \text { for } x \geq 0\) is \(f^{-1}(x)=\sqrt{x-1}\), and the verifications prove our solution is correct.

Key Concepts

One-to-One Functions

A function is called one-to-one if it assigns a unique output to every distinct input. This means that no two different inputs in the function's domain produce the same output.
Understanding one-to-one functions is essential when discussing inverse functions, because only one-to-one functions have inverses that are also functions.
To determine if a function is one-to-one, consider:

• The Horizontal Line Test: If every horizontal line crosses the graph of the function at most once, the function is one-to-one.
• Solving Algebraically: Ensure that if \( f(a) = f(b) \) implies \( a = b \) for all \( a, b \) in the domain, then the function is one-to-one.

Verifying Inverse Functions

To verify an inverse function, we must confirm that the operations "undo" each other, restoring the original input.
The process involves two equations:

• \( f(f^{-1}(x)) = x \)
• \( f^{-1}(f(x)) = x \)

For example, if \( f(x) = x^2 + 1 \) and its inverse \( f^{-1}(x) = \sqrt{x-1} \), verify by:

• Substitute \( f^{-1}(x) \) into \( f(x) \): Compute \( f(f^{-1}(x)) = (\sqrt{x-1})^2 + 1 = x \).
• Substitute \( f(x) \) into \( f^{-1}(x) \): Compute \( f^{-1}(f(x)) = \sqrt{(x^2+1) - 1} = x \).

If both are satisfied, the verification is complete, confirming that \( f^{-1}(x) \) indeed serves as the inverse of \( f(x) \).

Solving Equations for Inverses

Finding the inverse of a function involves solving an equation so that you express the original input as the new output.
Here's how to solve for inverses:

• Start with the equation \( y = f(x) \).
• Swap \( x \) and \( y \): This suggests that outputs become the new inputs, giving \( x = f(y) \).
• Solve for \( y \): Isolate \( y \) in terms of \( x \). For example, if \( x = y^2 + 1 \), subtract 1 and take the square root, resulting in \( y = \sqrt{x-1} \).

The expression attained by solving for \( y \) is your inverse, represented as \( f^{-1}(x) \). It is crucial for applications that the function is one-to-one to ensure this result is valid.