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Question
Find the values of x for which f(x) = 2x3 – 15x2 – 144x – 7 is
- Strictly increasing
- strictly decreasing
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Solution
f(x) = 2x3 – 15x2 – 144x – 7
∴ f'(x) = `"d"/("d"x)(2x^3 - 15x^2 - 144x - 7)`
= 2 × 3x2 – 15 × 2x – 144 × 1 – 0
= 6x2 – 30x – 144
= 6(x2 – 5x – 24)
(a) f is strictly increasing if f'(x) > 0
i.e., if 6(x2 – 5x – 24) > 0
i.e., if x2 – 5x –24 > 0
i.e., if x2 – 5x > 24
i.e., if `x^2 - 5x + (25)/(4) > 24 + (25)/(4)`
i.e., if `(x - 5/2)^2 > (121)/(4)`
i.e., if `x - (5)/(2) > (11)/(2) or x - (5)/(2) < - (11)/(2)`
i.e., if x > 8 or x < – 3
∴ f is strictly increasing if x < – 3 or x > 8.
(b) f is strictly decreasing if f''(x) < 0
i.e., if 6(x2 – 5x – 24) < 0
i.e., if x2 – 5x –24 < 0
i.e., if x2 – 5x < 24
i.e., if `x^2 - 5x + (25)/(4) < 24 + (25)/(4)`
i.e., if `(x - 5/2)^2 < (121)/(4)`
i.e., if `x - (5)/(2) < (11)/(2) or x - (5)/(2) > - (11)/(2)`
i.e., if `-(11)/(2) + (5)/(2) < x - (5)/(2) + (5)/(2) < (11)/(2) + (5)/(2)`
i.e., if – 3 < x < 8
∴ f is strictly decreasing if – 3 < x < 8.
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