Example 11 - Chapter 6 Class 12 Application of Derivatives

Last updated at May 29, 2023 by Teachoo

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

Example 11 - Chapter 6 Class 12 Application of Derivatives - Part 2

Example 11 - Chapter 6 Class 12 Application of Derivatives - Part 3

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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

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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