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Question
Find the values of x for which the function f(x) = x3 – 12x2 – 144x + 13 (a) increasing (b) decreasing
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Solution
f(x) = x3 – 12x2 – 144x + 13
∴ f'(x) = `d/dx(x^3 - 12x^2 - 144x + 13)`
= 3x2 – 12 x 2x – 144 x 1 + 0
= 3x2 – 24x – 144
= 3(x2 – 8x – 48)
(a) if is increasing if f'(x) ≥ 0
i.e. if 3(x2 – 8x – 48) ≥ 0
i.e. if x2 – 8x – 48 ≥ 0
i.e. if x2 – 8x ≥ 48
i.e. if x2 – 8x + 16 ≥ 48 + 16
i.e. if (x – 4)2 ≥ 64
i.e. if x – 4 ≥ 8 or x – 4 ≤ – 8
i.e. if x ≥ 12 or x ≤ – 4
∴ f is increasing if x ≤ – 4 or x ≥ 12,
i.e. x ∈ `( - oo, - 4] ∪ [12, oo)`.
(b) f is decreasing if f'(x) ≤ 0
i.e. if 3(x2 – 8x – 48) ≤ 0
i.e. if x2 – 8x – 48 ≤ 0
i.e. if x2 – 8x ≤ 48
i.e. if x2 – 8x + 16 ≤ 48 + 16
i.e. if (x – 4)2 ≤ 64
i.e. if – 8 ≤ x – 4 ≤ 8
i.e. if – 4 ≤ x ≤ 12
∴ f is decreasing if – 4 ≤ x ≤ 12, i.e. x ∈[– 4, 12].
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