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

Ex 6.5,3 - Chapter 6 Class 12 Application of Derivatives - Part 4
Ex 6.5,3 - Chapter 6 Class 12 Application of Derivatives - Part 5

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Ex 6.5, 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: (ii) 𝑔(π‘₯)=π‘₯3 –3π‘₯𝑔(π‘₯)=π‘₯3 –3π‘₯ Finding g’(𝒙) g’(π‘₯)=𝑑(π‘₯^3 βˆ’ 3π‘₯)/𝑑π‘₯ g’(π‘₯)=3π‘₯^2βˆ’3 Putting g’(𝒙)=𝟎 3π‘₯^2βˆ’3=0 3π‘₯^2=3 π‘₯^2=3/3 π‘₯^2=1 π‘₯=Β±1 So, x = 1 & x = –1 Finding g’’(𝒙) g’(π‘₯)=3π‘₯^2βˆ’3 g’’(π‘₯)=𝑑(3π‘₯^2βˆ’3)/𝑑π‘₯ = 6π‘₯βˆ’0 = 6π‘₯ Putting 𝒙=𝟏 in g’’(x) g’’(1)=6(1)= 6 > 0 Thus, g’’(π‘₯)>0 when π‘₯=1 β‡’ π‘₯=1 is point of local minima & g(π‘₯) is minimum at π‘₯=1 Local minimum value g(π‘₯)=π‘₯^3βˆ’3π‘₯ g(1)=(1)^3βˆ’3(1) =1βˆ’3 =βˆ’πŸ Putting 𝒙=βˆ’πŸ in g’’(x) g’’(βˆ’1)=6(βˆ’1)= –6 < 0 Thus, g’’(π‘₯)<0 when π‘₯=βˆ’1 β‡’ π‘₯=βˆ’1 is point of local maxima & g(π‘₯) is maximum at π‘₯=βˆ’1 Local minimum value g(π‘₯)=π‘₯^3βˆ’3π‘₯ g(βˆ’1)=(βˆ’1)^3βˆ’3(βˆ’1) =βˆ’1+3 =𝟐

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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.