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Question
Find the values of x for which the following functions are strictly increasing:
f(x) = 3 + 3x – 3x2 + x3
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Solution
f(x) = 3 + 3x – 3x2 + x3
∴ f'(x) = `d/dx(3 + 3x - 3x^2 + x^3)`
= 0 + 3 × 1 – 3 × 2x + 3x2
= 3 – 6x + 3x2
= 3(x2 – 2x + 1)
f is strictly increasing if f'(x) > 0
i.e. if 3(x2 – 2x + 1) > 0
i.e. if x2 – 2x + 1 > 0
i.e. if (x – 1)2 > 0
This is possible if x ∈ R and x ≠ 1
i.e. x ∈ R – { 1 }
∴ f is strictly increasing if x ∈ R – { 1 }.
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