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Find the values of x for which f(x) = 2x3 – 15x2 – 144x – 7 is Strictly increasing strictly decreasing

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Question

Find the values of x for which f(x) = 2x3 – 15x2 – 144x – 7 is

  1. Strictly increasing
  2. strictly decreasing
Sum
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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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Chapter 2: Applications of Derivatives - Exercise 2.4 [Page 90]

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SCERT Maharashtra Mathematics and Statistics (Arts and Science) [English] 12 Standard HSC
Chapter 2.2 Applications of Derivatives
Long Answers III | Q 5

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