Question

Verify the Rolle's theorem for the function $f\left(x\right)=(x-1)(x-2)(x-3),x\in [1,4]$

Answer 1

Now, Rolle's theorem states that, for a differentiable function, between two consecutive same valued points, there must be at least one stationary point in between i.e.

The function $f\left(x\right)=(x-1)(x-2)(x-3)\forall x\in [1,4]$ has same value at $x=1,x=2,x=3$ i.e. it is zero.

Hence, it can be divided into 2 intervals $x\in [1,2]$ and $x\in [2,3]$.

$f^{\prime }\left(c\right)=0$ where $c\in [a,b]\forall f\left(a\right)=f\left(b\right)$.

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

Now, put $f^{\prime }\left(x\right)=0$.

$\therefore (x-1)(x-2)+(x-1)(x-3)+(x-2)(x-3)=0$

$\therefore 3x^2-12x+11=0$

$\therefore x=\frac{12\pm \sqrt{{12}^2-4\times 3\times 11}}{2\times 3}$

$\therefore x=\frac{12\pm \sqrt{12}}{6}$

$\therefore x=2\pm \frac{1}{\sqrt{3}}$

$\therefore x=2+\frac{1}{\sqrt{3}}\in [2,3]$ and $x=2-\frac{1}{\sqrt{3}}\in [1,2]$

Hence proved.