Question: If $f(x) = \sqrt{4x - 9}$, then $f^{-1}(x) = ________$.
1. **State the problem:** Find the inverse function $f^{-1}(x)$ of the function $f(x) = \sqrt{4x - 9}$.
2. **Recall the definition of inverse function:** The inverse function $f^{-1}(x)$ satisfies $f(f^{-1}(x)) = x$ and $f^{-1}(f(x)) = x$. To find $f^{-1}(x)$, we start by replacing $f(x)$ with $y$:
$$y = \sqrt{4x - 9}$$
3. **Solve for $x$ in terms of $y$:
Square both sides to eliminate the square root:**
$$y^2 = 4x - 9$$
4. **Isolate $x$:
$$4x = y^2 + 9$$
$$x = \frac{y^2 + 9}{4}$$
5. **Swap $x$ and $y$ to express the inverse function:**
$$y = f^{-1}(x) = \frac{x^2 + 9}{4}$$
6. **Check the domain and range:**
Since $f(x) = \sqrt{4x - 9}$, the domain of $f$ is $x \geq \frac{9}{4}$ to keep the expression under the square root non-negative. The range of $f$ is $y \geq 0$.
Therefore, the domain of $f^{-1}$ is $x \geq 0$.
**Final answer:**
$$f^{-1}(x) = \frac{x^2 + 9}{4}$$