Problem 27 - College Algebra

Question

The functions 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 $$f\left(f^{-1}(x)\right)=x \text { and } f^{-1}(f(x))=x$$ $$ f(x)=\frac{2 x+1}{x-3} $$

Step-by-Step Solution

Answer

The inverse function of \(f(x) = \frac{2x + 1}{x - 3}\) is \(f^{-1}(x) = \frac{3x + 1}{x - 2}\). The verification process confirms the results, demonstrating both \(f(f^{-1}(x)) = x\) and \(f^{-1}(f(x)) = x.\)

Step 1: Find Inverse Function

Replace \(f(x)\) with \(y\), so the original function becomes \(y = \frac{2x + 1}{x - 3}\). Now, swap \(x\) and \(y\), generating \(x = \frac{2y + 1}{y - 3}\). Solve for \(y\) to find the inverse function: Multiply both sides by \(y - 3\) yields \(x(y - 3) = 2y + 1\), which simplifies to \(xy - 3x = 2y + 1\). Rearrange terms to group all the terms with 'y' on one side: \(xy - 2y = 3x + 1\). Factor \(y\) out and divide to derive \(y = \frac{3x + 1}{x - 2}\), therefore, \(f^{-1}(x) = \frac{3x + 1}{x - 2}\).

Step 2: Verification

To verify, we need to show that \(f(f^{-1}(x)) = x\) and \(f^{-1}(f(x)) = x\). Computing \(f(f^{-1}(x))\), we substitute \(f^{-1}(x)\) into \(f(x)\), this gives \(f\left(\frac{3x + 1}{x - 2}\right) = \frac{2\left(\frac{3x + 1}{x - 2}\right) + 1}{\frac{3x + 1}{x - 2} - 3} = x\), which verifies \(f(f^{-1}(x)) = x\). Now compute \(f^{-1}(f(x))\) by substituting \(f(x)\) into \(f^{-1}(x)\), this gives \(f^{-1}\left(\frac{2x + 1}{x - 3}\right) = \frac{3 \left(\frac{2x + 1}{x - 3}\right) + 1}{\frac{2x + 1}{x - 3} - 2} = x\), which verifies \(f^{-1}(f(x)) = x\) as well.

Key Concepts

Finding Inverse Functions

The process of finding inverse functions is akin to finding a pathway back from the output to the input of the original function. To discover an inverse function, commonly denoted as \(f^{-1}(x)\), we follow a series of steps designed to reverse the operations carried out by the original function.

Initially, we start by replacing the function notation \(f(x)\) with \(y\), which helps us to visualize and manipulate the equation more easily. After this replacement, we swap the roles of \(x\) and \(y\), treating the inputs as outputs and vice versa. The challenging part often lies in solving this new equation for \(y\), which, upon completion, yields the formula for the inverse function.

Here's an example approach in bullet points:

• Begin with the original function \(f(x)\) and replace it with \(y\).
• Exchange \(x\) with \(y\) in the equation.
• Solve for the new \(y\), which represents \(f^{-1}(x)\).

Solving for \(y\) can involve various algebraic techniques such as factoring, distributing, or isolating the variable. The ultimate goal is to have a clear expression for \(y\) that can effectively 'undo' the action of the original function.

Verification of Inverse Functions

After deducing the formula for an inverse function, it's crucial to ensure that the function truly reverses the actions of the original. This step of verification safeguards against any algebraic mistakes made during the finding process. There are two key equations we use for this confirmation:

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

By applying these equations, we can check that inserting the inverse function back into the original function—or vice versa—yields the identity function, symbolically speaking, it returns us back to our starting point. This bilateral verification confirms that both functions are indeed inverses of each other.

To carry out this method:

• Substitute \(f^{-1}(x)\) into \(f(x)\), and simplify the expression to check if the result is \(x\).
• Conversely, substitute \(f(x)\) into \(f^{-1}(x)\), and once again, simplify to check for \(x\).

If both conditions hold true, we have successfully validated the inverse function.

One-to-One Functions

To even begin considering inverses, we must ensure that our function is 'one-to-one'. A one-to-one function, also known as an injective function, is one where every element in the function's domain maps to a distinct, unique element in its range. This exclusivity is what allows a function to have an inverse; it guarantees that each output is associated with only one input.

One way to visually determine if a function is one-to-one is through the 'Horizontal Line Test'. If no horizontal line intersects the graph of the function more than once, the function is one-to-one. Algebraically, we can also check for every pair of distinct elements \(a\) and \(b\) in the domain such that \(f(a) eq f(b)\).

Without the one-to-one property, finding an inverse function is not possible because the function would not provide a clear, unambiguous reversal process. Hence, before calculating an inverse, we always verify that the function meets this crucial criterion.