Let `f:(-oo,2] to (-oo,4]` be a function defined by `f(x)=4x-x^(2)`. Then, `f^(-1)(x)` is

Options

A. `2- sqrt(4-x)`
B. `2+sqrt(4-x)`
C. `-2+ sqrt(4-x)`
D. `-2-sqrt(4-x)`

Answer

To find the inverse of the function \( f(x) = 4x - x^2 \) defined on the interval \( (-\infty, 2] \) and mapping to \( (-\infty, 4] \), we will follow these steps:

### Step 1: Set the function equal to \( y \)
We start by expressing the function in terms of \( y \):
\[ y = f(x) = 4x - x^2 \]

### Step 2: Rearrange the equation
Rearranging the equation gives us:
\[ x^2 - 4x + y = 0 \]

### Step 3: Use the quadratic formula
This is a quadratic equation in \( x \). We can apply the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1, b = -4, c = y \):
\[ x = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot y}}{2 \cdot 1} \]
\[ x = \frac{4 \pm \sqrt{16 - 4y}}{2} \]
\[ x = \frac{4 \pm 2\sqrt{4 - y}}{2} \]
\[ x = 2 \pm \sqrt{4 - y} \]

### Step 4: Determine the correct sign
Since the domain of \( f \) is \( (-\infty, 2] \), we need to choose the negative sign to ensure \( x \) remains in the correct range:
\[ x = 2 - \sqrt{4 - y} \]

### Step 5: Express the inverse function
Thus, we can express the inverse function \( f^{-1}(y) \) as:
\[ f^{-1}(y) = 2 - \sqrt{4 - y} \]

### Step 6: Replace \( y \) with \( x \)
Finally, we can replace \( y \) with \( x \) to write the inverse function in standard form:
\[ f^{-1}(x) = 2 - \sqrt{4 - x} \]

### Conclusion
The inverse function is:
\[ f^{-1}(x) = 2 - \sqrt{4 - x} \]