Show that the function $f (x) = x^{3} - 3 x^{2} + 6 x - 100$ is increasing on $R .$

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Solution

Given: $f (x) = x^{3} - 3 x^{2} + 6 x - 100$

$\Rightarrow f^{'} (x) = 3 x^{2} - 6 x + 6$

$= 3 (x^{2} - 2 x + 2)$

$= 3 (x^{2} - 2 x + 1 + 1)$

$= 3 {(x - 1)}^{2} + 3$

$∴ f^{'} (x) > 0$ , for all real values of $x$

Hence, $f (x)$ is increaseing on $R$ .

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