The problem consists of three parts related to a function $f(x)$ that satisfies a specific functional equation:

Given Information:

$f(2x) = a^{2 - 4x^2}$, where $a > 0$ and $a \neq 1$.

Tasks:

1. Find the explicit form of $f(x)$.
2. If $a = 2$, find the domain of the function $y = \sqrt{f(x) - \frac{1}{4}}$**.
3. Discuss the domain of $f(x)$.

1. Solving for $f(x)$:

We have the functional equation: $f(2x) = a^{2 - 4x^2}$ Let’s perform a substitution for $x = \frac{u}{2}$ so that $f(u) = a^{2 - u^2}$. Therefore, the function $f(x)$ is given by: $f(x) = a^{2 - x^2}$ Thus, the general form of $f(x)$ is: $f(x) = a^{2 - x^2}$

2. Finding the Domain of $y = \sqrt{f(x) - \frac{1}{4}}$ when $a = 2$:

When $a = 2$, the function $f(x)$ becomes: $f(x) = 2^{2 - x^2}$ Now, we need to find the domain of the function: $y = \sqrt{2^{2 - x^2} - \frac{1}{4}}$ For the square root to be defined, the expression inside must be non-negative: $2^{2 - x^2} - \frac{1}{4} \geq 0$ This simplifies to: $2^{2 - x^2} \geq \frac{1}{4}$ Taking the natural logarithm on both sides, we obtain: $2 - x^2 \geq -2$ Solving for $x$: $x^2 \leq 4$ Thus, $x$ must satisfy: $-2 \leq x \leq 2$ So, the domain of $y$ is $x \in [-2, 2]$.

3. Discuss the Domain of $f(x)$:

The function $f(x) = a^{2 - x^2}$, for $a > 0$, is always positive for all real $x$, since any power of a positive number remains positive. Hence, the domain of $f(x)$ is: $\text{Domain of } f(x): (-\infty, \infty)$

Would you like more details or have any specific questions about these solutions?

Here are some related questions to consider:

1. What happens to $f(x)$ as $x \to \infty$?
2. How does the value of $a$ affect the shape of $f(x)$?
3. What is the range of the function $f(x)$ for a specific value of $a$?
4. How would you find the inverse of the function $f(x)$?
5. Can the domain of $y$ change if $a$ is not equal to 2?

Tip: Always check the base $a$ when dealing with exponential functions to understand how it affects the function’s growth or decay.