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Question
The value of c in Rolle's theorem when
f (x) = 2x3 − 5x2 − 4x + 3, x ∈ [1/3, 3] is
Options
2
\[- \frac{1}{3}\]
−2
\[\frac{2}{3}\]
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Solution
2
Given: \[f\left( x \right) = 2 x^3 - 5 x^2 - 4x + 3\] Differentiating the given function with respect to x, we get
\[f'\left( x \right) = 6 x^2 - 10x - 4\]
\[ \Rightarrow f'\left( c \right) = 6 c^2 - 10c - 4\]
\[ \therefore f'\left( c \right) = 0 \]
\[ \Rightarrow 3 c^2 - 5c - 2 = 0\]
\[ \Rightarrow 3 c^2 - 6c + c - 2 = 0\]
\[ \Rightarrow 3c\left( c - 2 \right) + c - 2 = 0\]
\[ \Rightarrow \left( 3c + 1 \right)\left( c - 2 \right) = 0\]
\[ \Rightarrow c = 2, \frac{- 1}{3}\]
\[ \therefore c = 2 \in \left( \frac{1}{3}, 3 \right)\]
Thus,\[c = 2 \in \left( \frac{1}{3}, 3 \right)\] for which Rolle's theorem holds.
Hence, the required value of c is 2 .
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