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  4. The smallest value of the polynomial x3 - 18x2 + 96x in [0,9] is

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Q. The smallest value of the polynomial $x^3 - 18x^2 + 96x$ in $[0,9]$ is

Application of Derivatives

Solution:

$f(x) = x^3 - 18x^2 + 96x$
$\Rightarrow f'(x) = 3x^2 - 36x + 96$
$\therefore f'(x) = 0$
$\Rightarrow x^2 - 12x + 32 = 0$
$\Rightarrow x = 8$, $4$.
Now, $f(0) = 0$, $f(4) = 160$,
$f(8) = 128$, $f(9) = 135$
So, smallest value of $f(x)$ is $0$ at $x = 0$.