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Find the inverse of the function \( f(x)=\frac{x-3}{x+2} \) \( f^{-1}(x)=\frac{2 x+3}{1-x} \) \( f^{-1}(x)=\frac{x+3}{x-2} \) \( f^{-1}(x)=\frac{x+2}{x-3} \) \( f^{-1}(x)=\frac{2 x-3}{x-1} \)

Ask by Joseph Long. in the United States
Jan 22,2025

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Answer

The inverse of the function \( f(x) = \frac{x - 3}{x + 2} \) is \( f^{-1}(x) = \frac{2x + 3}{1 - x} \).

Solution

To find the inverse of the function \( f(x) = \frac{x - 3}{x + 2} \), follow these steps: 1. **Start with the original function:** \[ y = \frac{x - 3}{x + 2} \] 2. **Swap \( x \) and \( y \) to begin finding the inverse:** \[ x = \frac{y - 3}{y + 2} \] 3. **Solve for \( y \):** \[ x(y + 2) = y - 3 \] \[ xy + 2x = y - 3 \] \[ xy - y = -2x - 3 \] \[ y(x - 1) = -2x - 3 \] \[ y = \frac{-2x - 3}{x - 1} \] \[ y = \frac{2x + 3}{1 - x} \] 4. **Express the inverse function:** \[ f^{-1}(x) = \frac{2x + 3}{1 - x} \] **Therefore, the inverse of the function \( f(x) = \frac{x - 3}{x + 2} \) is:** \[ f^{-1}(x) = \frac{2x + 3}{1 - x} \] **Answer:** \( f^{-1}(x)=\frac{2 x+3}{1-x} \)

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Bonus Knowledge

To find the inverse of the function \( f(x)=\frac{x-3}{x+2} \), we start by setting \( y = \frac{x-3}{x+2} \). We then swap \( x \) and \( y \): \[ x = \frac{y - 3}{y + 2} \] Now, we solve for \( y \). First, multiply both sides by \( (y + 2) \): \[ x(y + 2) = y - 3 \] This simplifies to: \[ xy + 2x = y - 3 \] Rearranging gives: \[ xy - y = -3 - 2x \] Factoring out \( y \): \[ y(x - 1) = -3 - 2x \] Finally, solving for \( y \): \[ y = \frac{-3 - 2x}{x - 1} \] Thus, the inverse function is: \[ f^{-1}(x) = \frac{2x + 3}{1 - x} \] The correct choice is \( f^{-1}(x)=\frac{2x+3}{1-x} \). Now, let's learn a bit more about this function and its applications! Did you know that the function \( f(x) \) is a rational function, which means it can model real-world scenarios such as rates of change or proportions? This makes it handy in fields like economics or physics, where relationships often aren't linear! When working with inverses, one common mistake is forgetting to restrict the domain of the original function, which can lead to an incorrect inverse. Always double-check that the original function is one-to-one; otherwise, the inverse can't be properly defined!

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