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Question
It is given that the Rolle's theorem holds for the function f(x) = x3 + bx2 + cx, x \[\in\] at the point x = \[\frac{4}{3}\] , Find the values of b and c ?
Solution
As, the Rolle's theorem holds for the function f(x) = x3 + bx2 + cx, x \[\in\] [1, 2] at the point x = \[\frac{4}{3}\]
\[\text { So,} f\left( 1 \right) = f\left( 2 \right)\]
\[ \Rightarrow \left( 1 \right)^3 + b \left( 1 \right)^2 + c\left( 1 \right) = \left( 2 \right)^3 + b \left( 2 \right)^2 + c\left( 2 \right)\]
\[ \Rightarrow 1 + b + c = 8 + 4b + 2c\]
\[ \Rightarrow 3b + c + 7 = 0 . . . . . \left( i \right)\]
\[\text { And } f'\left( \frac{4}{3} \right) = 0\]
\[ \Rightarrow 3 \left( \frac{4}{3} \right)^2 + 2b\left( \frac{4}{3} \right) + c = 0 \left[ As, f'\left( x \right) = 3 x^2 + 2bx + c \right]\]
\[ \Rightarrow \frac{16}{3} + \frac{8b}{3} + c = 0\]
\[ \Rightarrow 8b + 3c + 16 = 0 . . . . . \left( ii \right)\]
\[\left( ii \right) - \left( i \right) \times 3, \text { we ge }\]
\[8b - 9b + 16 - 21 = 0\]
\[ \Rightarrow - b - 5 = 0\]
\[ \Rightarrow b = - 5\]
\[\text { Substituting b } = - 5 \text { in} \left( i \right), \text { we get }\]
\[3\left( - 5 \right) + c + 7 = 0\]
\[ \Rightarrow - 15 + c + 7 = 0\]
\[ \Rightarrow c = 8\]
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