The given problem asks for the original function $f(x)$, based on its inverse $f^{-1}(x) = \sqrt{\frac{1}{2}x}$.

To find the original function, recall that if $y = f^{-1}(x)$, then:

$f(y) = x$

Given $f^{-1}(x) = \sqrt{\frac{1}{2}x}$, let’s express this relationship as:

$y = \sqrt{\frac{1}{2}x}$

Squaring both sides to eliminate the square root:

$y^2 = \frac{1}{2}x$

Now, solving for $x$, we get:

$x = 2y^2$

Thus, $f(y) = 2y^2$, meaning the original function is $f(x) = 2x^2$.

Now, examining the answer choices:

• (A) $f(x) = \sqrt{\frac{1}{2}y}, y \geq 0$
• (B) $f(x) = \frac{1}{2}x^2, x \geq \frac{1}{2}$
• (C) $f(x) = 2x^2, x \geq 0$
• (D) $f(x) = 2x^2$

The correct answer is (D) $f(x) = 2x^2$, as this matches the original function without restrictions.

Would you like more details on inverse functions, or have any other questions?

Here are five related questions you could explore:

1. What are the general steps to find the inverse of a function?
2. How does the domain and range of a function relate to its inverse?
3. Why is it important to consider the domain $x \geq 0$ in some cases for functions involving square roots?
4. Can all functions have an inverse? What conditions must be met?
5. How does composition of functions verify that two functions are inverses of each other?

Tip: To check if two functions are inverses, compose them and see if the result is the identity function $f(f^{-1}(x)) = x$.