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Question
Find the values of x for which the following func- tions are strictly increasing : f(x) = x3 – 6x2 – 36x + 7
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Solution
f(x) = x3 – 6x2 – 36x + 7
∴ f'(x) = `d/dx(x^3 - 6x^2 - 36x + 7)`
= 3x2 – 6 x 2x – 36 x 1 + 0
= 3x2 – 12x – 36
= 3(x2 – 4x – 12)
f is strictly increasing if f'(x) > 0
i.e. if 3(x2 – 4x – 12) > 0
i.e. if x2 – 4x –12 > 0
i.e.if x2 – 4x > 12
i.e. if x2 – 4x + 4 > 12 + 4
i.e. if (x – 2)2 > 16
i.e. if x – 2 > 4 or x – 2 < – 4
i.e if x > 6 or x < – 2
∴ f is strictly increasing if x < – 2 or x > 6.
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