Question

Verify the Rolle's theorem for $f\left(x\right)=x(x-1{)}^{2}in[0,1]$

Answer 1

We have, $f\left(x\right)=x(x-1{)}^2,x\in [0,1]$.

(i) Since, $f\left(x\right)=x(x-1{)}^2$ is a polynomial function.
So, it is continuous in [0,1]

(ii) Now, $f^{\prime }\left(x\right)=x.\frac{d}{dx}(x-1{)}^2+(x-1{)}^2\frac{d}{dx}x=x.2(x-1)1+(x-1{)}^2=2x^2-2x+x^2+1-2x$
$=3x^2-4x+1$ which exists in (0,1)

So, f(x) is differentiable in (0,1)

(iii) Now, f(0)=0 and $f\left(1\right)=0\Rightarrow f\left(0\right)=f\left(1\right)$

f satisfies the above conditions of Rolle's theorem

Hence, by Rolle's theorem, $\exists c\in [0,1]suchthat{f}^{\prime }\left(c\right)=0\Rightarrow 3c^2-4c+1=0\Rightarrow 3c^2-3c-c+1=0\Rightarrow 3c(c-1)-1(c-1)=0\Rightarrow (3c-1)(c-1)=0\Rightarrow c=\frac{1}{3},1\Rightarrow \frac{1}{3}\in (0,1)$
Thus, we see that there exists a real number c in the open interval (0,1)
Hence, Rolle's theorem has been verified.