Question

Verify Rolle's theorem for the following function on the indicated interval  f (x) = (x² − 1) (x − 2) on [−1, 2] ?

Answer 1

Given :

\[f\left( x \right) = \left( x^2 - 1 \right)\left( x - 2 \right)\]

i.e. \[f\left( x \right) = x^3 - 2 x^2 - x + 2\]

We know that a polynomial function is everywhere derivable and hence continuous.
So, being a polynomial function, 

\[f\left( x \right)\] is continuous and derivable on \[\left[ - 1, 2 \right]\] .

Also, 

\[f\left( - 1 \right) = f\left( 2 \right) = 0\]

Thus, all the conditions of Rolle's theorem are satisfied.
Now, we have to show that there exists 

\[c \in \left( - 1, 2 \right)\] such that \[f'\left( c \right) = 0\] .

We have

\[f\left( x \right) = x^3 - x - 2 x^2 + 2\]

\[ \Rightarrow f'\left( x \right) = 3 x^2 - 4x - 1\]

\[ \therefore f'\left( x \right) = 0 \Rightarrow 3 x^2 - 4x - 1 = 0\]

\[ \Rightarrow x = \frac{- \left( - 4 \right) \pm \sqrt{\left( - 4 \right)^2 - 4 \times 3 \times \left( - 1 \right)}}{2 \times 3}\]

\[ \Rightarrow x = \frac{4 \pm \sqrt{16 + 12}}{6}\]

\[ \Rightarrow x = \frac{4 \pm \sqrt{28}}{6}\]

\[ \Rightarrow x = \frac{4 \pm 2\sqrt{7}}{6}\]

\[ \Rightarrow x = \frac{2 \pm \sqrt{7}}{3}\]

\[ \Rightarrow x = \frac{1}{3}\left( 2 - \sqrt{7} \right), \frac{1}{3}\left( 2 + \sqrt{7} \right)\]

Thus, 

\[c = \frac{1}{3}\left( 2 - \sqrt{7} \right), \frac{1}{3}\left( 2 + \sqrt{7} \right) \in \left( - 1, 2 \right) \text { such that } f'\left( c \right) = 0\]

Hence, Rolle's theorem is verified.