The image shows two functions $f(x)$ and $g(g)$, along with the task to express the inverse of $f$, i.e., $f^{-1}(x)$.

The function $f(x) = 5x^2 - 10x + 7$ is defined for $x \geq 1$, and you are asked to find the inverse of this function $f^{-1}(x)$.

Steps to Find the Inverse $f^{-1}(x)$:

1. Set $y = f(x)$: $y = 5x^2 - 10x + 7$

2. Solve for $x$ in terms of $y$: Rearrange the equation to express $x$ in terms of $y$. First, isolate the quadratic part: $5x^2 - 10x + 7 = y$ $5x^2 - 10x + (7 - y) = 0$

3. Apply the quadratic formula: For a quadratic equation $ax^2 + bx + c = 0$, the solutions for $x$ are: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Here, $a = 5$, $b = -10$, and $c = 7 - y$. Plug these into the quadratic formula: $x = \frac{-(-10) \pm \sqrt{(-10)^2 - 4(5)(7 - y)}}{2(5)}$ $x = \frac{10 \pm \sqrt{100 - 20(7 - y)}}{10}$ $x = \frac{10 \pm \sqrt{100 - 140 + 20y}}{10}$ $x = \frac{10 \pm \sqrt{20y - 40}}{10}$ $x = \frac{10 \pm 2\sqrt{5y - 10}}{10}$ $x = 1 \pm \frac{\sqrt{5y - 10}}{5}$

4. Choose the positive root: Since $x \geq 1$, we select the positive root: $x = 1 + \frac{\sqrt{5y - 10}}{5}$

5. Substitute $y$ with $x$ to find $f^{-1}(x)$: $f^{-1}(x) = 1 + \frac{\sqrt{5x - 10}}{5}$

Thus, the inverse function is: $f^{-1}(x) = 1 + \frac{\sqrt{5x - 10}}{5}$

Would you like further details on any step or an explanation of how this solution was derived?

Here are 5 related questions for deeper understanding:

1. How do you verify if two functions are inverses of each other?
2. What happens if we choose the negative root instead of the positive one when finding the inverse?
3. How can we determine the domain of an inverse function?
4. How does restricting the domain of a quadratic function help in finding its inverse?
5. Can every quadratic function have an inverse?

Tip: When dealing with quadratic functions, always check if the domain allows the function to be one-to-one before finding its inverse.