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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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