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Question
Verify Rolle's theorem for the following function on the indicated interval f (x) = (x − 1) (x − 2)2 on [1, 2] ?
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Solution
Given:
\[f\left( x \right) = \left( x - 1 \right) \left( x - 2 \right)^2\]
i.e. \[f\left( x \right) = x^3 + 4x - 4 x^2 - x^2 - 4 + 4x\]
\[f\left( x \right) = x^3 - 5 x^2 + 8x - 4\]
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 \[\left[ 1, 2 \right]\] .
Also,
\[f\left( 1 \right) = \left( 1 \right)^3 - 5 \left( 1 \right)^2 + 8\left( 1 \right) - 4 = 0\]
\[f\left( 2 \right) = \left( 2 \right)^3 - 5 \left( 2 \right)^2 + 8\left( 2 \right) - 4 = 0\]
\[ \therefore f\left( 1 \right) = f\left( 2 \right) = 0\]
Thus, all the conditions of Rolle's theorem are satisfied.
Now, we have to show that there exists
\[c \in \left( 1, 2 \right)\] such that
\[f'\left( c \right) = 0\].
We have
\[f\left( x \right) = x^3 + 8x - 5 x^2 - 4\]
\[ \Rightarrow f'\left( x \right) = 3 x^2 + 8 - 10x\]
\[ \therefore f'\left( x \right) = 0 \Rightarrow 3 x^2 - 10x + 8 = 0\]
\[ \Rightarrow 3 x^2 - 6x - 4x + 8 = 0\]
\[ \Rightarrow 3x\left( x - 2 \right) - 4\left( x - 2 \right) = 0\]
\[ \Rightarrow \left( x - 2 \right)\left( 3x - 4 \right)\]
\[ \Rightarrow x = 2, \frac{4}{3}\]
Thus,
\[c = \frac{4}{3} \in \left( 1, 2 \right) \text { such that } f'\left( c \right) = 0\] .
Hence, Rolle's theorem is verified.
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