Check sibling questions

Example 12 - Find intervals where f(x) = sin 3x is decreasing

Example 12 - Chapter 6 Class 12 Application of Derivatives - Part 2
Example 12 - Chapter 6 Class 12 Application of Derivatives - Part 3 Example 12 - Chapter 6 Class 12 Application of Derivatives - Part 4 Example 12 - Chapter 6 Class 12 Application of Derivatives - Part 5 Example 12 - Chapter 6 Class 12 Application of Derivatives - Part 6 Example 12 - Chapter 6 Class 12 Application of Derivatives - Part 7 Example 12 - Chapter 6 Class 12 Application of Derivatives - Part 8

This video is only available for Teachoo black users

Maths Crash Course - Live lectures + all videos + Real time Doubt solving!


Transcript

Example 12 Find intervals in which the function given by f (x) = sin 3x, x, ∈ [0, πœ‹/2] is (a) increasing (b) decreasing. f(π‘₯) = sin 3π‘₯ where π‘₯ ∈ [0 ,πœ‹/2] Finding f’(x) f’(π‘₯) = 𝑑(sin⁑3π‘₯ )/𝑑π‘₯ f’(π‘₯) = cos 3π‘₯ Γ— 3 f’(𝒙) = 3. cos 3𝒙 Putting f’(𝒙) = 0 3 cos 3π‘₯ = 0 cos 3π‘₯ = 0 We know that cos ΞΈ = 0 When ΞΈ = πœ‹/2 & 3πœ‹/2 So, for cos 3𝒙 = 0 3π‘₯ = πœ‹/2 & 3π‘₯ = 3πœ‹/2 π‘₯ = πœ‹/(2 Γ—3) & π‘₯ = 3πœ‹/(2 Γ— 3) 𝒙 = 𝝅/πŸ” & 𝒙 = 𝝅/𝟐 Since π‘₯ = πœ‹/6 ∈ [𝟎 ,𝝅/𝟐] & π‘₯ = πœ‹/2 ∈ [𝟎,𝝅/𝟐] ∴ Both values of π‘₯ are valid Plotting points on number line So, point π‘₯ = πœ‹/6 divides the interval into two disjoint intervals [0 ,πœ‹/6) and (πœ‹/6, πœ‹/2] Checking sign of f’(𝒙) f’(π‘₯) = 3. cos 3π‘₯ Case 1: For 𝒙 ∈ (𝟎 ,𝝅/πŸ”) 0<π‘₯<πœ‹/6 3 Γ— 0<3π‘₯<3πœ‹/6 𝟎<πŸ‘π’™<𝝅/𝟐 So when π‘₯ ∈ (0 ,πœ‹/6), then 3π‘₯ ∈ (0 , πœ‹/2) We know that cos 𝜽>𝟎 for 𝜽 ∈ (𝟎 , 𝝅/𝟐) cos 3x >0 for 3x ∈ (0 , πœ‹/2) cos 3x >0 for x ∈ (0 , πœ‹/6) 3 cos 3x >0 for x ∈ (0 , πœ‹/6) 𝒇′(𝒙)>𝟎 for x ∈ (0 , πœ‹/6) Since f’(0) = 3 and f’(𝝅/πŸ”) = 0 Therefore, f’(x) β‰₯ 0 for π‘₯ ∈ [0 , πœ‹/6] Thus, f(x) is increasing for π‘₯ ∈ [0 , πœ‹/6] Case 2: For 𝒙 ∈ (𝝅/πŸ”, 𝝅/𝟐) πœ‹/6<π‘₯<πœ‹/2 3 Γ— πœ‹/6<3π‘₯<3πœ‹/2 𝝅/𝟐<πŸ‘π’™<πŸ‘π…/𝟐 So when π‘₯ ∈(πœ‹/6 , πœ‹/2), then 3π‘₯ ∈ (πœ‹/2 , 3πœ‹/2) We know that, cos πœƒ<0 for πœƒ ∈ (πœ‹/2 , 3πœ‹/2) cos 3π‘₯<0 for 3π‘₯ ∈ (πœ‹/2 , 3πœ‹/2) cos 3π‘₯<0 for π‘₯ ∈ (πœ‹/6 , πœ‹/2) 3 cos 3π‘₯<0 for π‘₯ ∈ (πœ‹/6 , πœ‹/2) fβ€˜(x) <𝟎 for π‘₯ ∈ (πœ‹/6 , πœ‹/2) Since f’(𝝅/πŸ”) = 0 and f’(𝝅/𝟐) = 0 Therefore f’(x) ≀ 0 for π‘₯ ∈ [πœ‹/6,πœ‹/2] Thus, f(x) is decreasing for π‘₯ ∈ [πœ‹/6,πœ‹/2] (As cos πœƒ is negative in 2nd and 3rd quadrant) Thus, f(x) is increasing for 𝒙 ∈ [𝟎 , 𝝅/πŸ”] & f(x) is decreasing for 𝒙 ∈ [𝝅/πŸ” , 𝝅/𝟐]

Ask a doubt (live)
Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.