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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 ] ? - Mathematics

#### 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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