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# The Least Value of the Function F(X) = X 3 − 18 X 2 + 96 X in the Interval [0,9] is (A) 126 (B) 135 (C) 160 (D) 0 - Mathematics

#### Question

The least value of the function f(x) = $x3 - 18x2 + 96x$ in the interval [0,9] is _____________ .

• 126

• 135

• 160

• 0

#### Solution

0

$\text { Given }: f\left( x \right) = x^3 - 18 x^2 + 96x$

$\Rightarrow f'\left( x \right) = 3 x^2 - 36x + 96$

$\text { For a local maxima or a local minima, we must have }$

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

$\Rightarrow 3 x^2 - 36x + 96 = 0$

$\Rightarrow x^2 - 12x + 32 = 0$

$\Rightarrow \left( x - 4 \right)\left( x - 8 \right) = 0$

$\Rightarrow x = 4, 8$

$\text { So,}$

$f\left( 8 \right) = \left( 8 \right)^3 - 18 \left( 8 \right)^2 + 96\left( 8 \right) = 512 - 1152 + 768 = 128$

$f\left( 4 \right) = \left( 4 \right)^3 - 18 \left( 4 \right)^2 + 96\left( 4 \right) = 64 - 288 + 384 = 160$

$f\left( 0 \right) = \left( 0 \right)^3 - 18 \left( 0 \right)^2 + 96\left( 0 \right) = 0$

$f\left( 9 \right) = \left( 9 \right)^3 - 18 \left( 9 \right)^2 + 96\left( 9 \right) = 729 - 1458 + 864 = 135$

$\text { Hence, 0 is the minimum value in the range } \left[ 0, 9 \right] .$

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