Verify Rolles theorem for the function: `f(x)=x(x+3)e^(-x//2)` on `[-3,\ 0]` .

Explanation

To verify Rolle's Theorem for the function \( f(x) = x(x+3)e^{-\frac{x}{2}} \) on the interval \([-3, 0]\), we will follow the steps outlined in the theorem.

### Step 1: Check Continuity on the Closed Interval \([-3, 0]\)

**Explanation:**
Rolle's Theorem states that if \( f(x) \) is continuous on the closed interval \([a, b]\), then we can proceed with the theorem.

The function \( f(x) = x(x+3)e^{-\frac{x}{2}} \) is a product of three functions:
1. \( x \) (a polynomial, hence continuous everywhere),
2. \( x + 3 \) (also a polynomial, hence continuous everywhere),
3. \( e^{-\frac{x}{2}} \) (an exponential function, hence continuous everywhere).

Since all these functions are continuous, their product \( f(x) \) is also continuous on \([-3, 0]\).

### Step 2: Check Differentiability on the Open Interval \((-3, 0)\)

**Explanation:**
Next, we need to check if \( f(x) \) is differentiable on the open interval \((-3, 0)\).

Since \( f(x) \) is composed of polynomials and an exponential function, which are both differentiable everywhere, \( f(x) \) is differentiable on \((-3, 0)\).

### Step 3: Check if \( f(a) = f(b) \)

**Explanation:**
We need to check if \( f(-3) = f(0) \).

Calculating \( f(-3) \):
\[ f(-3) = (-3)(-3 + 3)e^{-\frac{-3}{2}} = (-3)(0)e^{\frac{3}{2}} = 0 \]

Calculating \( f(0) \):
\[ f(0) = (0)(0 + 3)e^{-\frac{0}{2}} = (0)(3)e^{0} = 0 \]

Since \( f(-3) = 0 \) and \( f(0) = 0 \), we have \( f(-3) = f(0) \).

### Step 4: Find \( c \) such that \( f'(c) = 0 \)

**Explanation:**
Now we need to find the derivative \( f'(x) \) and solve for \( c \) in the interval \((-3, 0)\) such that \( f'(c) = 0 \).

Using the product rule:
Let \( u = x(x+3) \) and \( v = e^{-\frac{x}{2}} \).

Then,
\[ f'(x) = u'v + uv' \]
Calculating \( u' \):
\[ u = x^2 + 3x \implies u' = 2x + 3 \]
Calculating \( v' \):
\[ v = e^{-\frac{x}{2}} \implies v' = e^{-\frac{x}{2}} \cdot \left(-\frac{1}{2}\right) \]

Now substituting back:
\[ f'(x) = (2x + 3)e^{-\frac{x}{2}} + (x^2 + 3x)\left(-\frac{1}{2}e^{-\frac{x}{2}}\right) \]
Factoring out \( e^{-\frac{x}{2}} \):
\[ f'(x) = e^{-\frac{x}{2}} \left( (2x + 3) - \frac{1}{2}(x^2 + 3x) \right) \]
Simplifying:
\[ f'(x) = e^{-\frac{x}{2}} \left( 2x + 3 - \frac{x^2}{2} - \frac{3x}{2} \right) \]
\[ = e^{-\frac{x}{2}} \left( -\frac{x^2}{2} + \frac{x}{2} + 3 \right) \]
Setting \( f'(x) = 0 \):
\[ -\frac{x^2}{2} + \frac{x}{2} + 3 = 0 \]
Multiplying through by -2:
\[ x^2 - x - 6 = 0 \]
Factoring:
\[ (x - 3)(x + 2) = 0 \]
Thus, \( x = 3 \) or \( x = -2 \).

Since we are looking for \( c \) in the interval \((-3, 0)\), we take \( c = -2 \).

### Conclusion

Since all conditions of Rolle's Theorem are satisfied, we conclude that there exists a \( c \) in the interval \((-3, 0)\) such that \( f'(c) = 0 \). In this case, \( c = -2 \).