Example 11 - Chapter 6 Class 12 Application of Derivatives

Last updated at Dec. 16, 2024 by

  Example 11 - Find intervals in which f(x) is strictly - Examples - Examples

part 2 - Example 11 - Examples - Serial order wise - Chapter 6 Class 12 Application of Derivatives
part 3 - Example 11 - Examples - Serial order wise - Chapter 6 Class 12 Application of Derivatives

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Example 11 Find the intervals in which the function f given by f (𝑥)=4𝑥3−6𝑥2 –72𝑥+30 is (a) strictly increasing (b) strictly decreasing. f (𝑥)=4𝑥3−6𝑥2 –72𝑥+30 Calculating f’(x) f (𝑥)=4𝑥3−6𝑥2–72𝑥+30 𝑓′(𝑥)=12𝑥2−12𝑥 –72𝑥 𝑓′(𝑥) =12(𝑥2−𝑥 – 6) 𝑓′(𝑥) =12(𝑥2−3𝑥+2𝑥 –6) 𝑓′(𝑥) = 12(𝑥(𝑥 − 3) + 2(𝑥 − 3)) 𝒇′(𝒙)=𝟏𝟐(𝒙+𝟐)(𝒙 –𝟑) Putting f’(x) = 0 12(𝑥+2)(𝑥 –3)=0 (𝑥+2)(𝑥 –3)=0 So, x = −2 and x = 3 Plotting points on number line Hence, f is strictly increasing in (−∞ ,−𝟐) & (𝟑 ,∞) f is strictly decreasing in (−𝟐, 𝟑)

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo