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Question
Find the values of x for which the following functions are strictly increasing : f(x) = 2x3 – 3x2 – 12x + 6
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Solution
f(x) = 2x3 – 3x2 – 12x + 6
∴ f'(x) = `d/dx(2x^3 - 3x^2 - 12x + 6)`
= 2 x 3x2 – 3 x 2x – 12 x 1 + 0
= 6x2 – 6x – 12
= 6(x2 – x – 2)
f is strictly increasing if f'(x) > 0
i.e. if 6(x2 – x – 2) > 0
i.e. if x2 – x – 2 > 0
i.e. if x2 – x > 2
i.e. if `x^2 - x + (1)/(4) > 2 + (1)/(4)`
i.e. if `(x - 1/2)^2 > (9)/(4)`
i.e. if `x - (1)/(2) > (3)/(2) or x - (1)/(2) < - (3)/(2)`
i.e. if x > 2 or x < – 1
∴ f is strictly increasing if x < – 1 or x > 2.
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