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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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