Misc 14 - Chapter 6 Class 12 Application of Derivatives

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Misc 14 Find the absolute maximum and minimum values of the function f given by f (𝑥) = cos2 𝑥 + sin⁡𝑥, 𝑥 ∈ [0, 𝜋 ] f(𝑥)=cos^2 𝑥+sin 𝑥 , 𝑥 ∈ [0 , 𝜋] Finding f’(𝒙) f(𝑥)=cos^2 𝑥+sin 𝑥 f’(𝑥)= 𝑑(cos^2⁡〖𝑥 + sin⁡𝑥 〗 )/𝑑𝑥 = 2cos 𝑥. 𝑑(cos 𝑥)/𝑑𝑥 + cos 𝑥 = 2cos 𝑥(−sin 𝑥)+cos⁡𝑥 = cos 𝑥 (−2sin 𝑥+1) Putting f’(𝒙) = 0 cos 𝑥 (−2 sin⁡〖𝑥+1〗 )=0 𝑥 = 𝜋/6 & (𝜋 )/2 are Critical points. cos 𝑥 = 0 cos 𝑥 = 0 cos 𝑥 = cos 𝜋/2 𝑥 = 𝜋/2 – 2 sin 𝑥 + 1 = 0 – 2 sin 𝑥 = –1 sin 𝑥 = (−1)/(−2) sin 𝑥 = 1/2 sin 𝑥 = sin 𝜋/6 𝑥 = 𝜋/6 Since our interval is 𝑥 ∈ [0, 𝜋 ] Critical points are 𝑥=0, 𝜋/6 , 𝜋/2 , 𝜋 Hence Absolute maximum value = 𝟓/𝟒 & Absolute minimum value = 1