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Question
Verify Rolle's theorem for the following function on the indicated interval f(x) = x2 − 4x + 3 on [1, 3] ?
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Solution
\[f\left( x \right) = x^2 - 4x + 3\]
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 \[\left[ 1, 3 \right]\] .
Also,
\[f\left( 1 \right) = \left( 1 \right)^2 - 4\left( 1 \right) + 3 = 1 - 4 + 3 = 0\]
\[f\left( 3 \right) = \left( 3 \right)^2 - 4\left( 3 \right) + 3 = 9 - 12 + 3 = 0\]
\[ \therefore f\left( 1 \right) = f\left( 3 \right) = 0\]
Thus, all the conditions of Rolle's theorem are satisfied.
Now, we have to show that there exists \[c \in \left( 1, 3 \right)\] such that \[f'\left( c \right) = 0\] .
We have
\[f\left( x \right) = x^2 - 4x + 3\]
\[ \Rightarrow f'\left( x \right) = 2x - 4\]
\[ \therefore f'\left( x \right) = 0 \Rightarrow 2x - 4 = 0 \Rightarrow x = 2\]
Thus,
\[c = 2 \in \left( 1, 3 \right) \text { such that } f'\left( c \right) = 0\]
Hence, Rolle's theorem is verified.
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