# 18) Prove that the function given by $f (x) = x^3 - 3x^2 + 3x - 100$ is increasing in R.

$f (x) = x^3 - 3x^2 + 3x - 100$
$f^{'}(x) = 3x^2 - 6x + 3$
$= 3(x^2 - 2x + 1) = 3(x-1)^2$
$f^{'}(x) = 3(x-1)^2$
We can clearly see that for any value of x in R $f^{'}(x) > 0$
Hence, $f (x) = x^3 - 3x^2 + 3x - 100$  is an increasing function in R