Browse Questions

# Verify the Rolle's theorem for the function $f(x)=x(x+3)e^{\large-x/2}\;in\;[-3,0]$

Toolbox:
• If $f[a,b]\rightarrow R$ is continuous on $[a,b]$ and differentiable on $(a,b)$ such that $f(a)=f(b)$,then there exists some c in $(a,b)$ such that $f'(c)=0$
Step 1:
$f(x)=x(x+3)e^{-\Large\frac{x}{2}}$ in $[-3,0]$
Since a polynomial function and an exponential function are everywhere continuous and differentiable.Therefore $f(x)$ being product of there two,is continuous on $[-3,0]$ and also differentiable on $(-3,0)$.
$f(-3)=-3(-3+3)e^{\large\frac{3}{2}}=0$
$f(0)=0$
Therefore $f(-3)=f(0)=0$
Thus $f(x)$ satisfied all the three conditions of Rolle's theorem on $[-3,0]$
Now $f(x)=x(x+3)e^{\Large\frac{-x}{2}}$
(Apply product rule)
$\Rightarrow f'(x)=(2x+3)e^{\Large\frac{-x}{2}}+(x^2+3x)(-\large\frac{1}{2})e^{-\Large\frac{x}{2}}$
Step 2:
On simplifying we get,
$f'(x)=e^{-\Large\frac{x}{2}}\bigg[\large\frac{-x^2+x+6}{2}\bigg]$
Therefore $f'(x)=0$
$\Rightarrow e^{\Large\frac{-x}{2}}\bigg[\large\frac{-x^2+x+6}{2}\bigg]=$$0$
$\Rightarrow -x^2+x-6=0$
On factorising we get,
$(x-3)(x+2)=0$
$\Rightarrow x=3,-2$
Therefore $c=-2\in (-3,0)$.Such that $f'(c)=0$.
Hence Rolle's theorem is verified.