Let's break down the problem step by step.

1. Find the x-intercepts of the function $f(x) = -4\sqrt{x} + 4$:

The x-intercepts are found by setting $f(x) = 0$ and solving for $x$.

$f(x) = -4\sqrt{x} + 4 = 0$

Solving for $x$:

$-4\sqrt{x} + 4 = 0$ Move the constant to the other side: $-4\sqrt{x} = -4$ Divide both sides by -4: $\sqrt{x} = 1$ Square both sides: $x = 1$

Thus, the larger x-value is $x = 1$.

Now, let's look for any additional x-intercepts. However, since square roots only allow non-negative values, the smallest possible x-value is 0. But, substituting $x = 0$ into the function:

$f(0) = -4\sqrt{0} + 4 = 4$

This means $x = 0$ is not an x-intercept. Hence, the only x-intercept is $x = 1$.

2. Showing that $f'(x) = 0$ at some point between the two x-intercepts:

The derivative $f'(x)$ will help us check where the slope of the function is zero.

First, let's find $f'(x)$. Recall that the derivative of $\sqrt{x}$ is: $\frac{d}{dx} \left( \sqrt{x} \right) = \frac{1}{2\sqrt{x}}$

So, the derivative of $f(x) = -4\sqrt{x} + 4$ is:

$f'(x) = -4 \times \frac{1}{2\sqrt{x}} = -\frac{2}{\sqrt{x}}$

Now, set $f'(x) = 0$:

$-\frac{2}{\sqrt{x}} = 0$

This equation has no solution, as the expression cannot equal zero for any real $x$. Therefore, $f'(x) = 0$ does not occur at any point. The solution to $f'(x) = 0$ seems incorrect.

3. Correct Interpretation:

The question asks for a value of $x$ such that $f'(x) = 0$, but no such $x$ exists. This means the answer provided for that part is incorrect.

Do you want further details or have any questions on this problem?

Here are five related questions:

1. What happens to the function $f(x)$ for negative values of $x$?
2. How would the behavior of the function change if it were $f(x) = 4\sqrt{x} + 4$ instead?
3. Can a square root function have multiple x-intercepts?
4. What does the derivative $f'(x)$ represent in terms of the slope of the graph?
5. Why does $f'(x)$ not have a solution for $f'(x) = 0$?

Tip: When solving for the derivative of a function involving square roots, make sure to simplify carefully and verify whether the derivative can be zero within the domain of the original function.