If `f(x)=(x+2)/(x-1)=y` then

Options

A. `x=f(y)`
B. `f(1)=3`
C. `f(y)=2f(x)`
D. `f(y)=2+f(x)`

Answer

To solve the problem given \( f(x) = \frac{x+2}{x-1} = y \), we need to analyze the function and derive some properties from it. Let's go step by step.

### Step 1: Rewrite the function
We start with the function:
\[ f(x) = \frac{x+2}{x-1} \]
We can express \( y \) in terms of \( x \):
\[ y = \frac{x+2}{x-1} \]

### Step 2: Cross-multiply to eliminate the fraction
To eliminate the fraction, we can cross-multiply:
\[ y(x - 1) = x + 2 \]
This gives us:
\[ yx - y = x + 2 \]

### Step 3: Rearrange the equation
Now, we will rearrange the equation to isolate \( x \):
\[ yx - x = y + 2 \]
Factoring out \( x \) from the left side:
\[ x(y - 1) = y + 2 \]

### Step 4: Solve for \( x \)
Now we can solve for \( x \):
\[ x = \frac{y + 2}{y - 1} \]

### Step 5: Verify the function
We can verify that this is indeed the inverse function by substituting back into the original function:
If we substitute \( y = f(x) \) back into the function:
\[ f\left(\frac{y + 2}{y - 1}\right) = \frac{\frac{y + 2}{y - 1} + 2}{\frac{y + 2}{y - 1} - 1} \]
Simplifying this expression will show that it returns \( y \).

### Conclusion
Thus, we have derived that if \( f(x) = \frac{x+2}{x-1} \), then the inverse function is:
\[ f^{-1}(y) = \frac{y + 2}{y - 1} \]