Find the maximum value of `(3-x)^(5)(2+x)^(4)` when x lies between `-2` and 3 .

Answer

To find the maximum value of the function \( f(x) = (3-x)^5 (2+x)^4 \) for \( x \) in the interval \([-2, 3]\), we can follow these steps:

### Step 1: Define the function
We start with the function:
\[ f(x) = (3-x)^5 (2+x)^4 \]

### Step 2: Find the critical points
To find the critical points, we need to differentiate the function and set the derivative equal to zero. We will use the product rule for differentiation.

Let:
\[ u = (3-x)^5 \quad \text{and} \quad v = (2+x)^4 \]

Then, the derivative \( f'(x) \) using the product rule is:
\[ f'(x) = u'v + uv' \]

Calculating \( u' \) and \( v' \):
\[ u' = -5(3-x)^4 \quad \text{and} \quad v' = 4(2+x)^3 \]

Thus, we have:
\[ f'(x) = -5(3-x)^4(2+x)^4 + (3-x)^5 \cdot 4(2+x)^3 \]

### Step 3: Set the derivative to zero
Setting \( f'(x) = 0 \):
\[ -5(3-x)^4(2+x)^4 + 4(3-x)^5(2+x)^3 = 0 \]

Factoring out common terms:
\[ (3-x)^4(2+x)^3 \left( -5(2+x) + 4(3-x) \right) = 0 \]

This gives us two cases:
1. \( (3-x)^4 = 0 \) which gives \( x = 3 \)
2. \( (2+x)^3 = 0 \) which gives \( x = -2 \)
3. Solve \( -5(2+x) + 4(3-x) = 0 \):
\[ -5(2+x) + 12 - 4x = 0 \implies -5x + 12 - 10 = 0 \implies -5x + 2 = 0 \implies x = \frac{2}{5} \]

### Step 4: Evaluate the function at critical points and endpoints
Now we evaluate \( f(x) \) at the critical points \( x = -2, \frac{2}{5}, 3 \):
1. \( f(-2) = (3 - (-2))^5 (2 + (-2))^4 = 5^5 \cdot 0^4 = 0 \)
2. \( f(3) = (3 - 3)^5 (2 + 3)^4 = 0^5 \cdot 5^4 = 0 \)
3. \( f\left(\frac{2}{5}\right) = \left(3 - \frac{2}{5}\right)^5 \left(2 + \frac{2}{5}\right)^4 = \left(\frac{13}{5}\right)^5 \left(\frac{12}{5}\right)^4 \)

Calculating \( f\left(\frac{2}{5}\right) \):
\[ f\left(\frac{2}{5}\right) = \frac{13^5}{5^5} \cdot \frac{12^4}{5^4} = \frac{13^5 \cdot 12^4}{5^9} \]

### Step 5: Compare values
The maximum value of \( f(x) \) occurs at \( x = \frac{2}{5} \):
\[ \text{Maximum value} = \frac{13^5 \cdot 12^4}{5^9} \]

### Final Result
Thus, the maximum value of \( f(x) \) in the interval \([-2, 3]\) is:
\[ \frac{13^5 \cdot 12^4}{5^9} \]