Question

The function $f\left(x\right)=x(x+3)e^{-\left(1/2\right)x}$ satisfies all the conditions of Rolle's theorem in [–3, 0]. The value of c is

• A. -1
• B. -2
• C. -3

Answer 1

The correct option is B -2

$f\left(x\right)=x(x+3)e^{-\left(1/2\right)x}$
f'(x) = $(x+3)e^{-\left(1/2\right)x}+x.e^{-\left(1/2\right)x}-\frac{(x+3)e^{-\left(1/2\right)x}}{2}$
As f(x) satisfies all the conditions of rolle's theorem we'll get atleast a point in the interval [-3, 0] where f'(x) will vanish.
f'(x) = 0
$(x+3)e^{-\left(1/2\right)x}+x.e^{-\left(1/2\right)x}-\frac{(x+3)e^{-\left(1/2\right)x}}{2}$ = 0
= $e^{-\left(1/2\right)x}(2x+3-fracx(x+3)2)$
x = -2, 3
In the interval [-3, 0] it'll be x = -2 .