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Question
Verify Rolle's theorem for the following function on the indicated interval f (x) = x(x − 1)2 on [0, 1] ?
Solution
\[f\left( x \right) = x \left( x - 1 \right)^2\]
\[\Rightarrow f\left( x \right) = x\left( x^2 - 2x + 1 \right)\]
\[\therefore f\left( x \right) = \left( x^3 - 2 x^2 + x \right)\]
We know that a polynomial function is everywhere derivable and hence continuous.
So,
\[f\left( x \right)\] being a polynomial function is continuous and derivable on \[\left[ 0, 1 \right]\] .
Also,
\[f\left( 0 \right) = f\left( 1 \right) = 0\]
Thus, all the conditions of Rolle's theorem are satisfied.
Now, we have to show that there exists \[c \in \left( 0, 1 \right)\] such that \[f'\left( c \right) = 0\]
We have
\[f\left( x \right) = x^3 - 2 x^2 + x\]
\[ \Rightarrow f'\left( x \right) = 3 x^2 - 4x + 1\]
\[ \therefore f'\left( x \right) = 0 \Rightarrow 3 x^2 - 4x + 1 = 0\]
\[ \Rightarrow 3 x^2 - 3x - x + 1 = 0\]
\[ \Rightarrow 3x\left( x - 1 \right) - 1\left( x - 1 \right) = 0\]
\[ \Rightarrow \left( x - 1 \right) \left( 3x - 1 \right) = 0\]
\[ \Rightarrow x = 1, \frac{1}{3}\]
Thus,
\[c = \frac{1}{3} \in \left( 0, 1 \right) \text { such that }f'\left( c \right) = 0\]
Hence, Rolle's theorem is verified.
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