# 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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