Check sibling questions

Example 13 - Find intervals where f(x) = sin x + cos x is

Example 13 - Chapter 6 Class 12 Application of Derivatives - Part 2
Example 13 - Chapter 6 Class 12 Application of Derivatives - Part 3 Example 13 - Chapter 6 Class 12 Application of Derivatives - Part 4 Example 13 - Chapter 6 Class 12 Application of Derivatives - Part 5 Example 13 - Chapter 6 Class 12 Application of Derivatives - Part 6


Transcript

Example 13 Find the intervals in which the function f given by f (π‘₯)=sin⁑π‘₯+cos⁑π‘₯ , 0 ≀ π‘₯ ≀ 2πœ‹ is strictly increasing or strictly decreasing.f(π‘₯) = sin π‘₯ + cos π‘₯ Finding f’(𝒙) f’(π‘₯) = (𝑑 )/𝑑π‘₯ (sin π‘₯ + cos π‘₯) f’(π‘₯) = 𝑑(sin⁑π‘₯ )/𝑑π‘₯ + 𝑑(cos⁑π‘₯ )/𝑑π‘₯ f’(π‘₯) = "cos " π‘₯ + (βˆ’π‘ π‘–π‘›π‘₯) f’(𝒙) = 𝒄𝒐𝒔⁑𝒙 – π’”π’Šπ’β‘π’™ Putting f’(𝒙) = 0 cos π‘₯ βˆ’ sin π‘₯ = 0 cos 𝒙 = sin 𝒙 ∴π‘₯=𝝅/πŸ’ ,πŸ“π…/πŸ’ π‘Žπ‘  0" ≀ " π‘₯ ≀ 2πœ‹ Plotting points So, points π‘₯=πœ‹/4 ,5πœ‹/4 divides interval into 3 disjoint intervals [0 , πœ‹/4), (πœ‹/4,5πœ‹/4), (5πœ‹/4 , 2πœ‹] Checking sign of 𝒇^β€² (𝒙) 𝑓^β€² (π‘₯)" "=" cos " π‘₯" – sin " π‘₯ When 𝒙 ∈ [𝟎 , 𝝅/πŸ’) Let us find value of f’(x) at any value of π‘₯ lies between 0, πœ‹/4 Thus, f’(𝒙) > 0 for π‘₯ ∈ [0 , πœ‹/4) At 𝒙 = 0 f’(0) = cos 0 – sin 0 = 1 – 0 = 1 > 0 At 𝒙 = 𝝅/πŸ” ∈ (𝟎 , 𝝅/πŸ’) f’(πœ‹/6) = cos πœ‹/6 – sin πœ‹/6 = √3/2 – 1/2 = (√3 βˆ’ 1)/2 =(1.73 βˆ’ 1)/2=0.73/2 > 0 When 𝒙 ∈ (𝝅/πŸ’,πŸ“π…/πŸ’) As πœ‹/4 < x < 5πœ‹/4 Let us find value of f’(x) at any value of π‘₯ lies between πœ‹/4, 5πœ‹/4 Thus, f’(𝒙) < 0 for π‘₯ ∈ (πœ‹/4,5πœ‹/4) Let 𝒙 = 𝝅/𝟐 ∈ (𝝅/πŸ’,πŸ“π…/πŸ’) f’ (π‘₯) = cos π‘₯ – sin π‘₯ f’(πœ‹/2) = cos πœ‹/2 – sin πœ‹/2 = 0 – 1 = – 1 < 0 When 𝒙 ∈ (πŸ“π…/πŸ’ , πŸπ…] As 5πœ‹/4 < π‘₯ ≀ 2πœ‹ Let us find value of f’(x) at any value of π‘₯ lies between 5πœ‹/4, 2πœ‹ At 𝒙 = 2Ο€ f’(π‘₯) = cos π‘₯ – sin π‘₯ f’(2πœ‹) = cos 2πœ‹ – sin 2πœ‹ = 1βˆ’0 = 1 > 0 Hence, f’(x) > 0 for π‘₯ ∈ (5πœ‹/4 , 2πœ‹] When 𝒙 ∈ (πŸ“π…/πŸ’ , πŸπ…] As 5πœ‹/4 < π‘₯ ≀ 2πœ‹ Let us find value of f’(x) at any value of π‘₯ lies between 5πœ‹/4, 2πœ‹ At 𝒙 = 2Ο€ f’(π‘₯) = cos π‘₯ – sin π‘₯ f’(2πœ‹) = cos 2πœ‹ – sin 2πœ‹ = 1βˆ’0 = 1 > 0 Hence, f’(x) > 0 for π‘₯ ∈ (5πœ‹/4 , 2πœ‹] Thus, f is strictly increasing in intervals [𝟎 , 𝝅/πŸ’)& (πŸ“π…/πŸ’ , πŸπ…] f is strictly increasing in intervals (𝝅/πŸ’ , πŸ“π…/πŸ’)

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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 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.