Question

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

• A. $0$
• B. $-1$
• C. $-2$
• D. $-3$

Answer 1

The correct option is C $-2$

Given $f\left(x\right)=x(x+3)e^{-\left(1/2\right)x}$

$\therefore f^{\prime }\left(x\right)=(x^2+3x)e^{-\left(1/2\right)x}(-\frac{1}{2})+(2x+3)e^{-\left(1/2\right)x}$

$=-\frac{1}{2}{e}^{-\left(1/2\right)x}\{x^2-x-6\}$

Since $f\left(x\right)$ satisfies the rolle's theorem

$\therefore f^{\prime }\left(x\right)=0\Rightarrow -\frac{1}{2}{e}^{-\left(c/2\right)}(c^2-c-6)=0$

$\Rightarrow c=3,-2$

But $c=3\notin [3,0]$

$\therefore c=-2$