Let `f (x)= {{:( xe ^(ax)"," , x le 0),( x+ ax ^(2)-x ^(3)"," , x gt 0):}` where `a` is a positive constant . The interval in which f '(x) is increasing is `[(k)/(a ),(a)/(l)],` Then `k +l` is equal to ______

Answer

To solve the problem, we need to analyze the function \( f(x) \) given in two parts, depending on whether \( x \) is less than or equal to 0 or greater than 0. We will find the first derivative \( f'(x) \), then the second derivative \( f''(x) \), and determine the intervals where \( f'(x) \) is increasing.

### Step 1: Define the function
The function is defined as:
\[ f(x) = \begin{cases} x e^{ax} & \text{if } x \leq 0 \\ x + ax^2 - x^3 & \text{if } x > 0 \end{cases} \]

### Step 2: Find the first derivative \( f'(x) \)
For \( x \leq 0 \):
Using the product rule:
\[ f'(x) = e^{ax} + x \cdot a e^{ax} = e^{ax}(1 + ax) \]

For \( x > 0 \):
\[ f'(x) = 1 + 2ax - 3x^2 \]

### Step 3: Find the second derivative \( f''(x) \)
For \( x \leq 0 \):
Using the product rule:
\[ f''(x) = a e^{ax}(1 + ax) + e^{ax} \cdot a = e^{ax}(a(1 + ax) + a) = e^{ax}(a + a^2x) \]

For \( x > 0 \):
\[ f''(x) = 2a - 6x \]

### Step 4: Determine where \( f'(x) \) is increasing
To find where \( f'(x) \) is increasing, we need to check where \( f''(x) > 0 \).

#### For \( x \leq 0 \):
\[ e^{ax}(a + a^2x) > 0 \]
Since \( e^{ax} > 0 \) for all \( x \), we have:
\[ a + a^2x > 0 \implies x > -\frac{1}{a} \]

#### For \( x > 0 \):
\[ 2a - 6x > 0 \implies 6x < 2a \implies x < \frac{a}{3} \]

### Step 5: Combine the intervals
From the analysis:
1. For \( x \leq 0 \): \( x > -\frac{1}{a} \)
2. For \( x > 0 \): \( x < \frac{a}{3} \)

Thus, the combined interval where \( f'(x) \) is increasing is:
\[ \left[-\frac{2}{a}, \frac{a}{3}\right] \]

### Step 6: Identify \( k \) and \( l \)
We can compare this interval with the given form \( \left[\frac{k}{a}, \frac{a}{l}\right] \):
- From \( -\frac{2}{a} \), we have \( k = -2 \)
- From \( \frac{a}{3} \), we have \( l = 3 \)

### Step 7: Calculate \( k + l \)
\[ k + l = -2 + 3 = 1 \]

### Final Answer
Thus, the value of \( k + l \) is:
\[ \boxed{1} \]