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To Prove: f(x) is increasing

Given: f(x) = x³ - 3x² + 6x - 100

Concept Used: f(x) is increasing when f'(x) ≥ 0

Explanation:

f(x) = x³ - 3x² + 6x - 100

Differentiating both sides, we get,

⇒ f'(x) = 3x² - 6x + 6

⇒ f'(x) = 3(x² - 2x + 2)

⇒ f'(x) = 3[(x-1)² + 1] ≥ 0 for all x ∈ R

So, f(x) is increasing function for all x ∈ R.