Question

Verify Rolle's theorem for the following function on the indicated interval   f (x) = x(x − 4)² on the interval [0, 4] ?

Answer 1

Given function is 

\[f\left( x \right) = x \left( x - 4 \right)^2\] , which can be rewritten as \[f\left( x \right) = x^3 - 8 x^2 + 16x\] .

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 \[\left[ 0, 4 \right]\] .

Also,

\[f\left( 0 \right) = f\left( 4 \right) = 0\]

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

\[c \in \left[ 0, 4 \right]\] such that \[f'\left( c \right) = 0\] .

We have

\[f\left( x \right) = x^3 - 8 x^2 + 16x\]

\[ \Rightarrow f'\left( x \right) = 3 x^2 - 16x + 16\]

\[ \therefore f'\left( x \right) = 0 \]

\[ \Rightarrow 3 x^2 - 16x + 16 = 0\]

\[ \Rightarrow 3 x^2 - 12x - 4x + 16 = 0\]

\[ \Rightarrow 3x\left( x - 4 \right) - 4\left( x - 4 \right) = 0\]

\[ \Rightarrow \left( x - 4 \right)\left( 3x - 4 \right)\]

\[ \Rightarrow x = 4, \frac{4}{3}\]

Thus, 

\[c = \frac{4}{3} \in \left( 0, 4 \right) \text { such that } f'\left( c \right) = 0\] .

Hence, Rolle's theorem is verified.