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Question
Find the values of x for which the following functions are strictly decreasing:
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 × 3x2 – 3 × 2x – 12 × 1 + 0
= 6x2 – 6x – 12
= 6(x2 – x – 2)
f is strictly decreasing 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 `-(3)/(2) < x - (1)/(2) < (3)/(2)`
i.e. if `-(3)/(2) + (1)/(2) < x -(1)/(2) + (1)/(2) < (3)/(2) + (1)/(2)`
i.e. if – 1 < x < 2
∴ f is strictly decreasing if – 1 < x < 2
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