# Show that the fun

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.

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