Question

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

Answer 1

The given function is \[f\left( x \right) = x \left( x - 2 \right)^2\] , which can be rewritten as \[f\left( x \right) = x^3 - 4 x^2 + 4x\] .

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 \[\left[ 0, 2 \right]\] .

Also,

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

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

We have

\[f\left( x \right) = x^3 - 4 x^2 + 4x\]

\[ \Rightarrow f'\left( x \right) = 3 x^2 - 8x + 4\]

\[\text { When } f'\left( x \right) = 0 \]

\[ 3 x^2 - 8x + 4 = 0\]

\[ \Rightarrow 3 x^2 - 6x - 2x + 4 = 0\]

\[ \Rightarrow 3x\left( x - 2 \right) - 2\left( x - 2 \right) = 0\]

\[ \Rightarrow \left( x - 2 \right)\left( 3x - 2 \right)\]

\[ \Rightarrow x = 2, \frac{2}{3}\]

Thus, 

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

Hence, Rolle's theorem is verified.