To find the inverse of the function $f(x) = -x^2 + 4x - 4$ for $x \leq 2$, we need to follow these steps:

1. Express the function as $y = f(x)$:

$y = -x^2 + 4x - 4$

2. Solve for $x$ in terms of $y$:

We need to solve the quadratic equation for $x$ in terms of $y$.

$-x^2 + 4x - 4 = y$

To make it easier to solve, multiply both sides by -1:

$x^2 - 4x + 4 = -y$

This simplifies to:

$x^2 - 4x + (4 + y) = 0$

Now, this is a standard quadratic equation in the form $ax^2 + bx + c = 0$, where $a = 1$, $b = -4$, and $c = 4 + y$.

3. Use the quadratic formula to solve for $x$:

The quadratic formula is:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Substituting $a = 1$, $b = -4$, and $c = 4 + y$:

$x = \frac{4 \pm \sqrt{16 - 4(1)(4 + y)}}{2(1)}$

Simplify the expression under the square root:

$x = \frac{4 \pm \sqrt{16 - 16 - 4y}}{2}$

$x = \frac{4 \pm \sqrt{-4y}}{2}$

$x = \frac{4 \pm 2i\sqrt{y}}{2}$

However, because we are dealing with real values of $y$ and we need $x$ to be real (since $y = f(x)$ and we're only considering $x \leq 2$), there must be an issue with the sign or approach.

Let's rewrite the equation:

4. Refactor to avoid complex values:

Rewriting the original function in a completed square form:

$f(x) = -(x^2 - 4x + 4)$ $f(x) = -(x - 2)^2$

So, we have:

$y = -(x - 2)^2$

$-(x - 2)^2 = y$

Solving for $x$:

$(x - 2)^2 = -y$

Taking the square root on both sides:

$x - 2 = \pm \sqrt{-y}$

Since $\sqrt{-y}$ must be a non-negative value and the function is defined for real values, we take the negative square root due to the restriction $x \leq 2$:

$x - 2 = -\sqrt{-y}$

Finally, solve for $x$:

$x = 2 - \sqrt{-y}$

5. Express the inverse function:

Thus, the inverse function $f^{-1}(y)$ is:

$f^{-1}(y) = 2 - \sqrt{-y}, \text{ for } y \leq 0$

Summary:

The inverse function is:

$f^{-1}(y) = 2 - \sqrt{-y} \quad \text{for} \quad y \leq 0$

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Would you like more details or have any questions? Here are some related questions:

1. How can you verify if two functions are inverses of each other?
2. What is the domain and range of the original function $f(x)$?
3. How does the restriction $x \leq 2$ affect the inverse function?
4. What is the graphical interpretation of finding an inverse?
5. How would you find the inverse if the function was defined for $x > 2$?

Tip: Always ensure the domain of the original function aligns with the range of the inverse function.