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Question
The least value of the function f(x) = \[x3 - 18x2 + 96x\] in the interval [0,9] is _____________ .
Options
126
135
160
0
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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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