Question

Verify Rolle's theorem for the following function on the indicated interval f(x) = x² − 8x + 12 on [2, 6] ?

Answer 1

Given :

\[f\left( x \right) = x^2 - 8x + 12\]

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 \[\left[ 2, 6 \right]\] .

Also,

\[f\left( 2 \right) = \left( 2 \right)^2 - 8\left( 2 \right) + 12 = 4 - 16 + 12 = 0\]
\[f\left( 6 \right) = \left( 6 \right)^2 - 8\left( 6 \right) + 12 = 36 - 48 + 12 = 0\]
\[ \therefore f\left( 2 \right) = f\left( 6 \right) = 0\]

Thus, all the conditions of Rolle's theorem are satisfied.
Now, we have to show that there exists \[c \in \left( 2, 6 \right)\] such that \[f'\left( c \right) = 0\] .

We have

\[f\left( x \right) = x^2 - 8x + 12\]

\[ \Rightarrow f'\left( x \right) = 2x - 8\]

\[ \therefore f'\left( x \right) = 0 \Rightarrow 2x - 8 = 0 \Rightarrow x = 4\]

Thus, 

\[c = 4 \in \left( 2, 6 \right) \text { such that } f'\left( c \right) = 0\]

Hence, Rolle's theorem is verified.