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Ex 6.5,5 - Chapter 6 Class 12 Application of Derivatives - Part 3

Ex 6.5,5 - Chapter 6 Class 12 Application of Derivatives - Part 4
Ex 6.5,5 - Chapter 6 Class 12 Application of Derivatives - Part 5

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Ex 6.5, 5 Find the absolute maximum value and the absolute minimum value of the following functions in the given intervals: (ii) f (π‘₯) = sin⁑π‘₯ + cos⁑π‘₯ , π‘₯ ∈ [0, πœ‹ ] Finding f’(𝒙) f’(π‘₯)=𝑑(𝑠𝑖𝑛π‘₯ + π‘π‘œπ‘ π‘₯)/𝑑π‘₯ f’(π‘₯)=cos⁑〖π‘₯ βˆ’sin⁑π‘₯ γ€— Putting f’(𝒙) cos⁑〖π‘₯ βˆ’sin⁑π‘₯ γ€—= 0 cos⁑〖π‘₯=sin⁑π‘₯ γ€— 1 = sin⁑π‘₯/(cos⁑ π‘₯) 1 = tan π‘₯ tan π‘₯ = 1 We know that know tan ΞΈ = 1 at ΞΈ = πœ‹/4 ∴ π‘₯ = πœ‹/4 Since given interval π‘₯ ∈ [0 , πœ‹] Hence calculating f(π‘₯) at π‘₯=0 , πœ‹/4 ,πœ‹ Absolute Maximum value of f(π‘₯) is √𝟐 at 𝒙 = 𝝅/πŸ’ & Absolute Minimum value of f(π‘₯) is –1 at 𝒙 = Ο€

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.