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Find the Absolute Maximum and Minimum Values of a Function F Given by F ( X ) = 2 X 3 − 15 X 2 + 36 X + 1 on the Interval [ 1 , 5 ] ? - CBSE (Science) Class 12 - Mathematics

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Question

Find the absolute maximum and minimum values of a function f given by \[f(x) = 2 x^3 - 15 x^2 + 36x + 1 \text { on the interval }  [1, 5]\] ?

 

Solution

\[\text { Given }: f\left( x \right) = 2 x^3 - 15 x^2 + 36x + 1\]

\[ \Rightarrow f'\left( x \right) = 6 x^2 - 30x + 36\]

\[\text {For a local maximum or a local minimum, we have }\]

\[ f'\left( x \right) = 0\]

\[ \Rightarrow 6 x^2 - 30x + 36 = 0\]

\[ \Rightarrow x^2 - 5x + 6 = 0\]

\[ \Rightarrow \left( x - 3 \right)\left( x - 2 \right) = 0\]

\[ \Rightarrow x = 2 \text{ and }x = 3 \]

\[\text { Thus, the critical points of f are 1, 2, 3 and 5 } . \]

\[\text { Now,} \]

\[f\left( 1 \right) = 2 \left( 1 \right)^3 - 15 \left( 1 \right)^2 + 36\left( 1 \right) + 1 = 24\]

\[f\left( 2 \right) = 2 \left( 2 \right)^3 - 15 \left( 2 \right)^2 + 36\left( 2 \right) + 1 = 29\]

\[f\left( 3 \right) = 2 \left( 3 \right)^3 - 15 \left( 3 \right)^2 + 36\left( 3 \right) + 1 = 28\]

\[f\left( 5 \right) = 2 \left( 5 \right)^3 - 15 \left( 5 \right)^2 + 36\left( 5 \right) + 1 = 56\]

\[\text { Hence, the absolute maximum value when x = 5 is 56 and the absolute minimum value when x = 1 is 24 .} \]

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Solution Find the Absolute Maximum and Minimum Values of a Function F Given by F ( X ) = 2 X 3 − 15 X 2 + 36 X + 1 on the Interval [ 1 , 5 ] ? Concept: Graph of Maxima and Minima.
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