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