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Question
The function f(x) = \[2 x^3 - 15 x^2 + 36x + 4\] is maximum at x = ________________ .
Options
3
0
4
2
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Solution
2
\[\text { Given }: f\left( x \right) = 2 x^3 - 15 x^2 + 36x + 4\]
\[ \Rightarrow f'\left( x \right) = 6 x^2 - 30x + 36\]
\[\text { For a local maxima or a local minima, we must have } \]
\[f'\left( x \right) = 0\]
\[ \Rightarrow 6 x^2 - 30x + 36 = 0\]
\[ \Rightarrow x^2 - 5x + 6 = 0\]
\[ \Rightarrow \left( x - 2 \right)\left( x - 3 \right) = 0\]
\[ \Rightarrow x = 2, 3\]
\[\text { Now }, \]
\[f''\left( x \right) = 12x - 30\]
\[ \Rightarrow f''\left( 2 \right) = 24 - 30 = - 6 < 0\]
\[\text { So, x = 1 is a local maxima } . \]
\[\text { Also }, \]
\[f''\left( 3 \right) = 36 - 30 = 6 > 0\]
\[\text { So,x = 2 is a local maxima }.\]
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