Rolle's Theorem Engineering Maths, Btech first year

Rolle's Theorem

If $f(x)$ is
a) continuous in [a,b]
b) differentiable in (a,b)
c) $f(a)=f(b)$
then there exists at least one value $c\in (a,b)$ such that $f^{\prime }(c)=0$

Sample Problem

Verify Rolle's Theorem for $$f(x)=x(x-2)e^{\frac{3x}{4}}\text{ in }(0,2)$$

Solution

The function is continuous and differentiable in (0,2)
Here $a=0\text{ and }b=2$ $$f(a)=f(0)=0$$ $$f(b)=f(2)=2(2-2)e^{\frac{6}{4}}=0$$ $$\text{As }f(a)=f(b)$$ $\Rightarrow$ Rolle's Theorem is applicable. Now, $$f^{\prime }(x)=(x^2-2x)e^{\frac{3x}{4}}\times \frac{3}{4}+e^{\frac{3x}{4}}(2x-2)$$ According to Rolle's Theorem, $$f^{\prime }(c)=0$$ $$\Rightarrow (x^2-2x)e^{\frac{3x}{4}}\times \frac{3}{4}+e^{\frac{3x}{4}}(2x-2)$$ $$e^{\frac{3x}{4}}\ne 0$$ $$\frac{3}{4}(x^2-2x)+2x-2=0$$ $$3x^2-2x+8x-8=0$$ $$3x^2+6x-8=0$$ $$\Rightarrow x=-2,x=\frac{8}{6}$$ $$\text{Or }c=-2,c=\frac{8}{6}$$ $$\text{Since }c=-2\notin (0,2)$$ $$\Rightarrow \boxed{{\displaystyle c=\frac{8}{6}}}$$