Question

If Rolle's theorem is applicable for the function \(f(x) = x(x+3)e^{-x/2}\) on \([-3, 0]\), then the value of \(c\) is:

• A. \(3\)
• B. \(3 \text{ and } -2\)
• C. \(-2\)
• D. \(-1\)

Hint:

Remember the conditions for Rolle's theorem: continuity, differentiability, and f(a) = f(b).

Answer 1

The Correct Option is C

Step 1: Verify the conditions for Rolle's theorem.
The function f(x) = x(x+3)e^-x/2 is continuous on [-3, 0] and differentiable on (-3, 0) since it is a product of polynomial and exponential functions.
Also, f(-3) = (-3)(-3+3)e^-(-3)/2 = (-3)(0)e^3/2 = 0
and f(0) = 0(0+3)e^-0/2 = 0(3)e⁰ = 0.
Since f(-3) = f(0) = 0, Rolle's theorem is applicable.

Step 2: Find the derivative of f(x).
f(x) = (x² + 3x)e^-x/2
f'(x) = (2x + 3)e^-x/2 + (x² + 3x)e^-x/2(-1/2)
f'(x) = e^-x/2(2x + 3 - (x² + 3x)/2)
f'(x) = e^-x/2((4x + 6 - x² - 3x)/2)
f'(x) = (e^-x/2/2)(-x² + x + 6)

Step 3: Set f'(c) = 0 and solve for c.
By Rolle's theorem, there exists a c \(\in\) (-3, 0) such that f'(c) = 0.
(e^-c/2/2)(-c² + c + 6) = 0
Since e^-c/2 \(\neq\) 0, we have -c² + c + 6 = 0.
c² - c - 6 = 0
(c - 3)(c + 2) = 0
c = 3 or c = -2.
Since c \(\in\) (-3, 0), we have c = -2.
Therefore, the value of c is -2.