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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


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Ex 6.3, 3 Find the local maxima and local minima, if any, of the following functions. Find also the local maximum and the local minimum values, as the case may be: (iv) f (π‘₯)=sin⁑π‘₯ –cos⁑π‘₯, 0<π‘₯<2 πœ‹f (π‘₯)=sin⁑π‘₯ –cos⁑π‘₯, 0<π‘₯<2 πœ‹ Finding f’(𝒙) f’(π‘₯)=cos⁑π‘₯βˆ’(βˆ’sin⁑π‘₯ ) f’(π‘₯)=cos⁑π‘₯+sin⁑π‘₯ Putting f’(𝒙)=𝟎 cos⁑π‘₯+sin⁑π‘₯ = 0 cos⁑π‘₯=βˆ’sin⁑π‘₯ 1=(βˆ’sin⁑π‘₯)/cos⁑π‘₯ (βˆ’sin⁑π‘₯)/cos⁑π‘₯ =1 – tan π‘₯=1 tan π‘₯=βˆ’1 Since 0 < π‘₯ < 2Ο€ & tan π‘₯ is negative tan ΞΈ lies in either 2nd or 4th quadrant So, value of π‘₯ is π‘₯=3πœ‹/4 π‘œπ‘Ÿ 7πœ‹/4 Now finding f’’(π‘₯) f’’(π‘₯)=𝑑(cos⁑π‘₯ + sin⁑π‘₯ )/𝑑π‘₯ f’’(π‘₯)=βˆ’sin⁑π‘₯+cos⁑π‘₯ Putting 𝒙 = πŸ‘π…/πŸ’ f’’(3πœ‹/4)=βˆ’π‘ π‘–π‘›(3πœ‹/4)+π‘π‘œπ‘ (3πœ‹/4) =βˆ’π‘ π‘–π‘›(πœ‹βˆ’πœ‹/4)+π‘π‘œπ‘ (πœ‹βˆ’ πœ‹/4) =βˆ’π‘ π‘–π‘›(πœ‹/4)+(βˆ’π‘π‘œπ‘  πœ‹/4) =(βˆ’1)/√2βˆ’1/√2 =(βˆ’2)/√2 =βˆ’βˆš2 < 0 Hence f’’(π‘₯)<0 when π‘₯ = 3πœ‹/4 Thus π‘₯ = 3πœ‹/4 is point of local maxima ∴ f(π‘₯) is maximum value at π‘₯ = 3πœ‹/4 The local maximum value is f(π‘₯)=sin⁑π‘₯βˆ’cos⁑π‘₯ f(3πœ‹/4)=𝑠𝑖𝑛(3πœ‹/4)βˆ’π‘π‘œπ‘ (3πœ‹/4) =𝑠𝑖𝑛(πœ‹βˆ’πœ‹/4)βˆ’π‘π‘œπ‘ (πœ‹βˆ’πœ‹/4) =𝑠𝑖𝑛(πœ‹/4)βˆ’(βˆ’π‘π‘œπ‘  πœ‹/4) =𝑠𝑖𝑛 πœ‹/4+π‘π‘œπ‘  πœ‹/4 =1/√2+1/√2 =2/√2 =√2 Now, for 𝒙 = πŸ•π…/πŸ’ f’’(π‘₯)=βˆ’sin⁑π‘₯+cos⁑π‘₯ Putting π‘₯ = 7πœ‹/4 f’’(7πœ‹/4)=βˆ’sin⁑(7πœ‹/4)+cos⁑(7πœ‹/4) f’’(7πœ‹/4)=βˆ’π‘ π‘–π‘›(2πœ‹βˆ’πœ‹/4)+π‘π‘œπ‘ (2πœ‹βˆ’πœ‹/4) =βˆ’(βˆ’π‘ π‘–π‘›(πœ‹/4))+π‘π‘œπ‘ (πœ‹/4) =𝑠𝑖𝑛 πœ‹/4+π‘π‘œπ‘  πœ‹/4 =1/√2 + 1/√2 =2/√2 =√2 > 0 f’’(π‘₯)>0 when π‘₯ = 7πœ‹/4 Thus π‘₯ = 7πœ‹/4 is point of local minima f(π‘₯) has minimum value at π‘₯ = 7πœ‹/4 Local minimum value is f(π‘₯)=𝑠𝑖𝑛π‘₯βˆ’π‘π‘œπ‘ π‘₯ f(7πœ‹/4)=𝑠𝑖𝑛(7πœ‹/4)βˆ’π‘π‘œπ‘ (7πœ‹/4) =𝑠𝑖𝑛(2πœ‹βˆ’πœ‹/4)βˆ’π‘π‘œπ‘ (2πœ‹βˆ’πœ‹/4) =βˆ’π‘ π‘–π‘›(πœ‹/4)βˆ’π‘π‘œπ‘ (πœ‹/4) =(βˆ’1)/√2 βˆ’ 1/√2 =(βˆ’2)/√2 =βˆ’βˆš2 Thus, f(π‘₯) is maximum at x = πŸ‘π…/πŸ’ and maximum value is √𝟐 & f(π‘₯) is minimum at x = πŸ•π…/πŸ’ and maximum value is β€“βˆšπ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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.