Prove that the function f(x) = \(x^3 – 3x^2 + 3x – 100\) is increasing on R

Solution :

We have, f(x) = \(x^3 – 3x^2 + 3x – 100\)

\(\implies\)  f'(x) = \(3x^2 – 6x + 3\) = \(3(x – 1)^2\)

Now, x \(\in\) R \(\implies\)  \((x – 1)^2\)  \(\ge\)  0  \(\implies\)  f'(x)  \(\ge\) 0.

Thus, f'(x) \(\ge\) 0 for all x \(\in\) R.

Hence, f(x) is increasing on R.

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