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Question
Find the values of x for which the function f(x) = x3 – 6x2 – 36x + 7 is strictly increasing
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Solution
f(x) = x3 – 6x2 – 36x + 7
∴ f′(x) = 3x2 – 12x – 36
= 3(x2 – 4x – 12)
= 3(x – 6)(x + 2)
f(x) is strictly increasing, if f′(x) > 0
∴ 3(x – 6)(x + 2) > 0
∴ (x – 6)(x + 2) > 0
ab > 0 ⇔ a > 0 and b > 0 or a < 0 and b < 0
Either x – 6 > 0 and x + 2 > 0
or
x – 6 < 0 and x + 2 < 0
Case I: x – 6 > 0 and x + 2 > 0
∴ x > 6 and x > – 2
∴ x > 6
Case II: x – 6 < 0 and x + 2 < 0
∴ x < 6 and x < – 2
∴ x < – 2
Thus, f(x) is strictly increasing for x ∈ (−∞ −2) ∪ (6, ∞).
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