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