verify Rolle's theorem for the function `f(x)=x(x+3)e^(-x/2)` in `[-3,0]`

Explanation

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

### Step 1: Check the continuity of \( f(x) \) on \([-3, 0]\)

The function \( f(x) \) is a product of a polynomial \( x(x + 3) \) and an exponential function \( e^{-x/2} \). Both polynomial and exponential functions are continuous everywhere in their domains.

Since the product of continuous functions is also continuous, we conclude that \( f(x) \) is continuous on \([-3, 0]\).

### Step 2: Check the differentiability of \( f(x) \) on \([-3, 0]\)

Similarly, since both the polynomial and the exponential function are differentiable everywhere, their product \( f(x) \) is also differentiable on \([-3, 0]\).

### Step 3: Evaluate \( f(-3) \) and \( f(0) \)

Now, we will calculate \( f(-3) \) and \( f(0) \):

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

\[ f(0) = 0(0 + 3)e^{0} = 0 \cdot 3 \cdot 1 = 0 \]

Since \( f(-3) = f(0) = 0 \), the conditions of Rolle's theorem are satisfied.

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

Next, 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, we have:

\[ f(x) = x(x + 3)e^{-x/2} \]

Let \( u = x(x + 3) \) and \( v = e^{-x/2} \).

Then,

\[ f'(x) = u'v + uv' \]

Calculating \( u' \):

\[ u' = (x^2 + 3x)' = 2x + 3 \]

Calculating \( v' \):

\[ v' = \left(e^{-x/2}\right)' = -\frac{1}{2}e^{-x/2} \]

Now substituting back into the derivative:

\[ f'(x) = (2x + 3)e^{-x/2} + x(x + 3)\left(-\frac{1}{2}e^{-x/2}\right) \]

Factoring out \( e^{-x/2} \):

\[ f'(x) = e^{-x/2} \left( (2x + 3) - \frac{1}{2}x(x + 3) \right) \]

Setting \( f'(x) = 0 \):

\[ e^{-x/2} \left( (2x + 3) - \frac{1}{2}x^2 - \frac{3}{2}x \right) = 0 \]

Since \( e^{-x/2} \) is never zero, we focus on the quadratic equation:

\[ (2x + 3) - \frac{1}{2}x^2 - \frac{3}{2}x = 0 \]

Multiplying through by 2 to eliminate the fraction:

\[ 4x + 6 - x^2 - 3x = 0 \]

This simplifies to:

\[ -x^2 + x + 6 = 0 \]

Rearranging gives:

\[ x^2 - x - 6 = 0 \]

Factoring:

\[ (x - 3)(x + 2) = 0 \]

Thus, the solutions are:

\[ x = 3 \quad \text{and} \quad x = -2 \]

### Step 5: Verify that \( c \) lies in \([-3, 0]\)

The only solution that lies in the interval \([-3, 0]\) is \( c = -2 \).

### Conclusion

Since all conditions of Rolle's theorem are satisfied, we have verified that there exists a \( c \in [-3, 0] \) such that \( f'(c) = 0 \).

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