Misc 14 - Find absolute max, min values f(x) = cos2 x + sin x

Misc 14 - Chapter 6 Class 12 Application of Derivatives - Part 2
Misc 14 - Chapter 6 Class 12 Application of Derivatives - Part 3

  1. Chapter 6 Class 12 Application of Derivatives (Term 1)
  2. Serial order wise

Transcript

Misc 14 Find the absolute maximum and minimum values of the function f given by f (π‘₯) = cos2 π‘₯ + sin⁑π‘₯, π‘₯ ∈ [0, πœ‹]f(π‘₯)=cos^2 π‘₯+sin π‘₯ , π‘₯ ∈ [0 , πœ‹] Finding f’(𝒙) f’(π‘₯)= 𝑑(cos^2⁑〖π‘₯ + sin⁑π‘₯ γ€— )/𝑑π‘₯ = 2cos π‘₯. 𝑑(cos π‘₯)/𝑑π‘₯ + cos π‘₯ = 2cos π‘₯(βˆ’sin π‘₯)+cos⁑π‘₯ = cos 𝒙 (βˆ’πŸπ¬π’π§ 𝒙+𝟏) Putting f’(𝒙) = 0 cos π‘₯ (βˆ’2 sin⁑〖π‘₯+1γ€— )=0 π‘₯ = πœ‹/6 , 5πœ‹/6 & πœ‹/2 are Critical points. cos 𝒙 = 0 cos π‘₯ = 0 cos π‘₯ = cos πœ‹/2 𝒙 = 𝝅/𝟐 – 2 sin 𝒙 + 1 = 0 – 2 sin π‘₯ = –1 sin π‘₯ = (βˆ’1)/(βˆ’2) sin π‘₯ = 1/2 sin π‘₯ = sin πœ‹/6 𝒙 = 𝝅/πŸ” Also, 𝒙 = πœ‹ βˆ’πœ‹/6=πŸ“π…/πŸ” Since our interval is 𝒙 ∈ [0, πœ‹] Critical points are π‘₯=𝟎, πœ‹/6 , πœ‹/2 ,5πœ‹/6,𝝅 Hence Absolute maximum value = πŸ“/πŸ’ & Absolute minimum value = 1

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.