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

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Solution

$f (x) = x^{3} - 3 x^{2} + 3 x - 100$
$f^{'} (x) = 3 x^{2} - 6 x + 3$
$f^{'} (x) = 3 (x^{2} - 2 x + 1)$
$f^{'} (x) = 3 (x - 1)^{2} > 0$
Thus, $f^{'} (x)$ is always positive on R.
Hence, the given function is increasing on R.

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Prove that the function given by is increasing in R.

f (x) = x^{3} - 3 x^{2} + 3 x - 100

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

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

R .

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