Prove that the function given by $f (x) = x^{3} - 3 x^{2} + 3 x - 10$ is increasing in R.

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Solution

Given, $f (x) = x^{3} - 3 x^{2} + 3 x - 100$
$\Rightarrow f^{'} (x) = 3 x^{2} - 6 x + 3 = 3 (x^{2} - 2 x + 1) = 3 (x - 1)^{2}$
For any $x ϵ R, (x - 1)^{2} > 0$ since, a perfect square cannot be negative
Then, $f^{'} (x) > 0$ . Hence, the given function f is an increasing function on R.

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