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

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0 	ext { Given }: f\left( x ight) = x^3 - 18 x^2 + 96x \Rightarrow f'\left( x ight) = 3 x^2 - 36x + 96 	ext { For a local maxima or a local minima, we must have } f'\left( x ight) = 0 \Rightarrow 3 x^2 - 36x + 96 = 0 \Rightarrow x^2 - 12x + 32 = 0 \Rightarrow \left( x - 4 ight)\left( x - 8 ight) = 0 \Rightarrow x = 4, 8 	ext { So,} f\left( 8 ight) = \left( 8 ight)^3 - 18 \left( 8 ight)^2 + 96\left( 8 ight) = 512 - 1152 + 768 = 128 f\left( 4 ight) = \left( 4 ight)^3 - 18 \left( 4 ight)^2 + 96\left( 4 ight) = 64 - 288 + 384 = 160 f\left( 0 ight) = \left( 0 ight)^3 - 18 \left( 0 ight)^2 + 96\left( 0 ight) = 0 f\left( 9 ight) = \left( 9 ight)^3 - 18 \left( 9 ight)^2 + 96\left( 9 ight) = 729 - 1458 + 864 = 135 	ext { Hence, 0 is the minimum value in the range } \left[ 0, 9 ight] .

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