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Question
Let f(x) = 2x3\[-\] 3x2\[-\] 12x + 5 on [ 2, 4]. The relative maximum occurs at x = ______________ .
Options
-2
-1
2
4
Solution
2
\[\text { Given }: f\left( x \right) = 2 x^3 - 3 x^2 - 12x + 5\]
\[ \Rightarrow f'\left( x \right) = 6 x^2 - 6x - 12\]
\[\text { For a local maxima or a local minima, we must have }\]
\[f'\left( x \right) = 0\]
\[ \Rightarrow 6 x^2 - 6x - 12 = 0\]
\[ \Rightarrow x^2 - x - 2 = 0\]
\[ \Rightarrow \left( x - 2 \right)\left( x + 1 \right) = 0\]
\[ \Rightarrow x = 2, - 1\]
\[\text{ Now, } \]
\[f''\left( x \right) = 12x - 6\]
\[ \Rightarrow f''\left( - 1 \right) = - 12 - 6 = - 18 < 0\]
\[\text { So, x = 1 is a local maxima } . \]
\[\text { Also }, \]
\[f''\left( 2 \right) = 24 - 6 = 18 > 0\]
\[\text { So, x = 2 is a local minima } . \]
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