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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


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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 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.