Determine the critical points of the function $f (x) = 2 x^{3} - 15 x^{2} + 36 x - 3$ .

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Solution

$f (x) = 2 x^{3} - 15 x^{2} + 36 x - 3$
Extreme values occur at $f^{'} (x) = 0$
$f^{'} (x) = 6 x^{2} - 30 x + 36 = 0$
$x^{2} - 5 x + 6 = 0$
$x^{2} - 2 x - 3 x + 6 = 0$
$x (x - 2) - 3 (x - 2) = 0$
$x = 2, x = 3$

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