To find the greatest possible value of $k$ such that the function $f(x) = x^2 - 2x + 7$ is one-to-one and its inverse, follow these steps:

1. Determine the Vertex: The given function $f(x) = x^2 - 2x + 7$ is a quadratic function. It has a parabolic shape, and to make it one-to-one, we need to restrict its domain such that it is either increasing or decreasing. The vertex of the parabola will be the point where the function changes from increasing to decreasing or vice versa.

   The vertex form of a quadratic function $ax^2 + bx + c$ can be found using $x = -\frac{b}{2a}$. Here, $a = 1$, $b = -2$, and $c = 7$.

   So, the x-coordinate of the vertex is: $x = -\frac{-2}{2 \cdot 1} = 1$ At $x = 1$, the function is at its minimum point.

2. Greatest Possible Value of $k$: To ensure $f(x)$ is one-to-one, we need to restrict the domain to the left of the vertex or to the right. Given $x \leq k$, we should restrict the domain to $x \leq 1$ to keep the function decreasing (one-to-one).

   Thus, the greatest possible value of $k$ is $1$.

3. Find the Inverse Function: Given $f(x) = x^2 - 2x + 7$ with the restricted domain $x \leq 1$, we need to find the inverse.

   • Rewrite $y = f(x)$: $y = x^2 - 2x + 7$

   • Express this in the form $ax^2 + bx + c = 0$: $x^2 - 2x + (7 - y) = 0$

   • Solve for $x$ using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $x = \frac{2 \pm \sqrt{4 - 4(7 - y)}}{2}$ $x = \frac{2 \pm \sqrt{4 - 28 + 4y}}{2}$ $x = \frac{2 \pm \sqrt{4y - 24}}{2}$ $x = 1 \pm \sqrt{y - 6}$

   Since the domain is restricted to $x \leq 1$, we take the solution: $x = 1 - \sqrt{y - 6}$

   • Thus, the inverse function is: $f^{-1}(x) = 1 - \sqrt{x - 6}$

So, the greatest possible value of $k$ is $1$, and the inverse function is: $f^{-1}(x) = 1 - \sqrt{x - 6}$

Would you like more details or have any questions about this solution?

Here are 5 questions you might want to ask:

1. How do we verify that a function is one-to-one on a given interval?
2. Can you explain the process of completing the square for finding the vertex of a quadratic function?
3. How do we graph the original function and its inverse to visualize their relationship?
4. What are some common methods to restrict the domain of a function to make it one-to-one?
5. How does the quadratic formula help in finding the inverse of a quadratic function?

Tip: When finding the inverse of a quadratic function, always consider the domain restriction to ensure the function is one-to-one. This will help you identify the correct branch of the inverse.