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# F ( X ) = 3 X 4 − 8 X 3 + 12 X 2 − 48 X + 25 in [ 0 , 3] - Mathematics

#### Question

f(x) = 3x^4 - 8x^3 + 12x^2- 48x + 25 " in "[0,3] .

#### Solution

$\text { Given }: \hspace{0.167em} f\left( x \right) = 3 x^4 - 8 x^3 + 12 x^2 - 48x + 25$

$\Rightarrow f'\left( x \right) = 12 x^3 - 24 x^2 + 24x - 48$

$\text { For a local maximum or a local minimum, we must have }$

$f'\left( x \right) = 0$

$\Rightarrow 12 x^3 - 24 x^2 + 24x - 48 = 0$

$\Rightarrow x^3 - 2 x^2 + 2x - 4 = 0$

$\Rightarrow x^2 \left( x - 2 \right) + 2\left( x - 2 \right) = 0$

$\Rightarrow \left( x - 2 \right)\left( x^2 + 2 \right) = 0$

$\Rightarrow x - 2 = 0 or \left( x^2 + 2 \right) = 0$

$\Rightarrow x = 2$

$\text { No real root exists for } \left( x^2 + 2 \right) = 0 .$

$\text { Thus, the critical points of f are 0, 2 and 3 } .$

$\text { Now },$

$f\left( 0 \right) = 3 \left( 0 \right)^4 - 8 \left( 0 \right)^3 + 12 \left( 0 \right)^2 - 48\left( 0 \right) + 25 = 25$

$f\left( 2 \right) = 3 \left( 2 \right)^4 - 8 \left( 2 \right)^3 + 12 \left( 2 \right)^2 - 48\left( 2 \right) + 25 = - 39$

$f\left( 3 \right) = 3 \left( 3 \right)^4 - 8 \left( 3 \right)^3 + 12 \left( 3 \right)^2 - 48\left( 3 \right) + 25 = 16$

$\text { Hence, the absolute maximum value when x = 0 is 25 and the absolute minimum value when x = 2 is - 39 } .$

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