Question

If the function \(\mathrm{f}(x)=x(x+3) \mathrm{e}^{-\frac{x}{2}}\) satisfies all the conditions of Rolle's theorem in \([-3,0]\), then c is

• A. 0
• B. -1
• C. -2
• D. -3

Hint:

By Rolle's theorem, c is a point in the interval where f'(c) = 0.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
\(f(-3)=0=f(0)\), so Rolle's theorem applies and there is a \(c\in(-3,0)\) with \(f'(c)=0\).

Step 2: Differentiate
\(f(x)=(x^{2}+3x)e^{-x/2}\), so \(f'(x)=e^{-x/2}\left[(2x+3)-\dfrac{x^{2}+3x}{2}\right]\).

Step 3: Solve f'(x) = 0
\(4x+6-x^{2}-3x=0\) gives \(x^{2}-x-6=0\), so \(x=3\) or \(x=-2\).

Step 4: Choose c
Only \(x=-2\) lies in \((-3,0)\), so \(c=-2\), option (C).

Final Answer:
The value of \(c\) is \(-2\). \[ \boxed{\text{(C) }-2} \]