Verified Answer :
- Set the original function \(y = \frac{8+x}{3x-2}\).
- Swap \(x\) and \(y\) to define the inverse relationship: \(x = \frac{8+y}{3y-2}\).
- Solve the equation for \(y\) by rearranging terms: \(3xy - 2x = 8 + y \implies y(3x - 1) = 2x + 8\).
- State the final inverse function: \(f^{-1}(x) = \frac{2x+8}{3x-1}\).
- The final answer is \(\boxed{f^{-1}(x) = \frac{2x+8}{3x-1}}\).
Explanation
- Problem Analysis
We are given a linear fractional function \(f(x) = \frac{8+x}{3x-2}\). Our goal is to find its inverse function \(f^{-1}(x)\). For a function to have an inverse, it must be a bijection. A linear fractional transformation \(f(x) = \frac{ax+b}{cx+d}\) is invertible if the determinant \(ad-bc\) is non-zero. Here, \(a=1, b=8, c=3, d=-2\). The determinant is \(ad-bc = (1)(-2) - (8)(3) = -2 - 24 = -26\). Since the determinant is not zero, the function is indeed invertible. The domain of \(f(x)\) excludes \(x = \frac{2}{3}\) as it results in a zero denominator.
- Setting up the Inverse Equation
To find the inverse function \(f^{-1}(x)\), we start by setting \(y = f(x)\):
\[y = \frac{8+x}{3x-2}\]
To find the inverse, we swap the roles of \(x\) and \(y\), so that \(y\) represents the inverse function \(f^{-1}(x)\):
\[x = \frac{8+y}{3y-2}\]
- Solving for the Inverse Function
Now, we solve the equation \(x = \frac{8+y}{3y-2}\) for \(y\).
Multiply both sides by \((3y-2)\):
\[x(3y-2) = 8+y\]
Distribute \(x\) on the left side:
\[3xy - 2x = 8+y\]
Next, move all terms containing \(y\) to the left side and all other terms to the right side:
\[3xy - y = 2x + 8\]
Factor out \(y\) from the left side:
\[y(3x-1) = 2x + 8\]
Finally, divide by \((3x-1)\) to isolate \(y\):
\[y = \frac{2x+8}{3x-1}\]
Thus, the inverse function is \(f^{-1}(x) = \frac{2x+8}{3x-1}\).
Examples
The concept of an inverse function is essential in many real-life situations where you need to reverse a process. For example, if you have a currency conversion formula \(f(x) = ax + b\) that converts dollars to another currency, finding the inverse \(f^{-1}(x)\) allows you to easily convert that currency back into dollars. In our case, the function \(f(x) = \frac{8+x}{3x-2}\) could represent a complex transformation in signal processing or data encoding; knowing its inverse allows for decoding the original signal. By understanding how to algebraically derive inverse functions, you gain a powerful tool for reversing operations and solving problems in engineering, economics, and mathematics.