The function f(x)=x(x+3) e-x / 2 satisfies all the conditions of Rolle's Theorem on [-3,0] . The value of c which verifies Rolle's Theorem, is

Question

The function f(x)=x(x+3) e-x / 2 satisfies all the conditions of Rolle's Theorem on [-3,0] . The value of c which verifies Rolle's Theorem, is (A) 0 (B) -1 (C) -2 (D) (-3/2)

Tardigrade Admin · Accepted Answer

f'(x)=0 ⇒ (x+3) e-x / 2+x e-x / 2-(x / 2)(x+3) e-x / 2=0 ⇒ x2-x-6=0 ⇒ x=-2,3 ⇒ c=-2 ∈(-3,0)

Q. The function $f (x) = x (x + 3) e^{- x /2}$ satisfies all the conditions of Rolle's Theorem on $[- 3, 0] .$ The value of $c$ which verifies Rolle's Theorem, is

0

-1

-2

$\frac{- 3}{2}$

$f^{'} (x) = 0$
$\Rightarrow (x + 3) e^{- x /2} + x e^{- x /2} - (x /2) (x + 3) e^{- x /2} = 0$
$\Rightarrow x^{2} - x - 6 = 0$
$\Rightarrow x = - 2, 3$
$\Rightarrow c = - 2 \in (- 3, 0)$

Q. The function f(x)=x(x+3)e−x/2 satisfies all the conditions of Rolle's Theorem on [−3,0]. The value of c which verifies Rolle's Theorem, is

0

-1

-2

2−3​

f′(x)=0 ⇒(x+3)e−x/2+xe−x/2−(x/2)(x+3)e−x/2=0 ⇒x2−x−6=0 ⇒x=−2,3 ⇒c=−2∈(−3,0)

Q. The function $f (x) = x (x + 3) e^{- x /2}$ satisfies all the conditions of Rolle's Theorem on $[- 3, 0] .$ The value of $c$ which verifies Rolle's Theorem, is

$\frac{- 3}{2}$

$f^{'} (x) = 0$
$\Rightarrow (x + 3) e^{- x /2} + x e^{- x /2} - (x /2) (x + 3) e^{- x /2} = 0$
$\Rightarrow x^{2} - x - 6 = 0$
$\Rightarrow x = - 2, 3$
$\Rightarrow c = - 2 \in (- 3, 0)$