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Find the values of x for which the function f(x) = x3 – 6x2 – 36x + 7 is strictly increasing

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Question

Find the values of x for which the function f(x) = x3 – 6x2 – 36x + 7 is strictly increasing

Sum
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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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Chapter 2.2: Applications of Derivatives - Short Answers II

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