Find the interval in which `f(x) = xsqrt(4ax-x^2), (alt0)` is decreasing

Answer

To find the interval in which the function \( f(x) = x \sqrt{4ax - x^2} \) is decreasing, we will follow these steps:

### Step 1: Identify the domain of the function
The expression under the square root must be non-negative for \( f(x) \) to be defined. Therefore, we need:
\[ 4ax - x^2 \geq 0 \]
Rearranging gives:
\[ -x^2 + 4ax \geq 0 \quad \Rightarrow \quad x^2 - 4ax \leq 0 \]
Factoring out \( x \):
\[ x(x - 4a) \leq 0 \]
The critical points are \( x = 0 \) and \( x = 4a \). Thus, the function is defined in the interval:
\[ [0, 4a] \]

### Step 2: Differentiate the function
Next, we differentiate \( f(x) \) using the product rule:
\[ f(x) = x \cdot \sqrt{4ax - x^2} \]
Let \( u = x \) and \( v = \sqrt{4ax - x^2} \). The derivative \( f'(x) \) is given by:
\[ f'(x) = u'v + uv' \]
where \( u' = 1 \) and \( v' = \frac{1}{2\sqrt{4ax - x^2}}(4a - 2x) \).

So,
\[ f'(x) = \sqrt{4ax - x^2} + x \cdot \frac{1}{2\sqrt{4ax - x^2}}(4a - 2x) \]

### Step 3: Simplify the derivative
Now we simplify \( f'(x) \):
\[ f'(x) = \sqrt{4ax - x^2} + \frac{x(4a - 2x)}{2\sqrt{4ax - x^2}} \]
To combine these terms, we find a common denominator:
\[ f'(x) = \frac{2(4ax - x^2) + x(4a - 2x)}{2\sqrt{4ax - x^2}} \]
This simplifies to:
\[ f'(x) = \frac{8ax - 2x^2 + 4ax - 2x^2}{2\sqrt{4ax - x^2}} = \frac{12ax - 4x^2}{2\sqrt{4ax - x^2}} = \frac{4x(3a - x)}{\sqrt{4ax - x^2}} \]

### Step 4: Determine where the derivative is less than or equal to zero
To find where \( f(x) \) is decreasing, we set \( f'(x) \leq 0 \):
\[ \frac{4x(3a - x)}{\sqrt{4ax - x^2}} \leq 0 \]
Since the denominator \( \sqrt{4ax - x^2} \) is positive in the interval \( (0, 4a) \), we focus on the numerator:
\[ 4x(3a - x) \leq 0 \]
This gives us two critical points:
1. \( x = 0 \)
2. \( x = 3a \)

### Step 5: Analyze the intervals
We analyze the sign of \( 4x(3a - x) \) in the intervals:
- \( (-\infty, 0) \)
- \( (0, 3a) \)
- \( (3a, 4a) \)
- \( (4a, \infty) \)

1. For \( x < 0 \): \( 4x(3a - x) > 0 \)
2. For \( 0 < x < 3a \): \( 4x(3a - x) > 0 \)
3. For \( x = 3a \): \( 4x(3a - x) = 0 \)
4. For \( 3a < x < 4a \): \( 4x(3a - x) < 0 \)
5. For \( x = 4a \): \( 4x(3a - x) = 0 \)

Thus, \( f(x) \) is decreasing in the interval \( (3a, 4a) \).

### Final Answer
The function \( f(x) = x \sqrt{4ax - x^2} \) is decreasing in the interval:
\[ (3a, 4a) \]