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Find the values of x, for which the function f(x) = x3 + 12x2 + 36ЁЭСе + 6 is monotonically decreasing
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f(x) = x3 + 12x2 + 36ЁЭСе + 6
∴ f′(x) = 3x2 + 24x + 36
= 3(x2 + 8x + 12)
= 3(x + 2)(x + 6)
f(x) is monotonically decreasing, if f′(x) < 0
∴ 3(x + 2)(x + 6) < 0
∴ (x + 2)(x + 6) < 0
ab < 0 ⇔ a > 0 and b < 0 or a < 0 and b > 0
∴ Either x + 2 > 0 and x + 6 < 0
or
x + 2 < 0 and x + 6 > 0
Case I: x + 2 > 0 and x + 6 < 0
∴ x > – 2 and x < – 6,
which is not possible.
Case II: x + 2 < 0 and x + 6 > 0
∴ x < – 2 and x > – 6
Thus, f(x) is monotonically decreasing for x ∈ (– 6, – 2).
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