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Ex 6.5,3 - Chapter 6 Class 12 Application of Derivatives - Part 20

Ex 6.5,3 - Chapter 6 Class 12 Application of Derivatives - Part 21
Ex 6.5,3 - Chapter 6 Class 12 Application of Derivatives - Part 22
Ex 6.5,3 - Chapter 6 Class 12 Application of Derivatives - Part 23

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Transcript

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: (vi) g (π‘₯) = π‘₯/2 + 2/π‘₯ , π‘₯ > 0g (π‘₯) = π‘₯/2 + 2/π‘₯, π‘₯ > 0 Finding g’(𝒙) g’(π‘₯)=𝑑/𝑑π‘₯ (π‘₯/2+2/π‘₯) =𝑑/𝑑π‘₯ (π‘₯/2)+𝑑/𝑑π‘₯ (2π‘₯^(βˆ’1) ) =1/2βˆ’2π‘₯^(βˆ’2) =1/2βˆ’2/π‘₯^2 Putting g’(𝒙)=𝟎 1/2βˆ’2/π‘₯^2 =0 (π‘₯^2βˆ’ 4)/(2π‘₯^2 )=0 π‘₯^2βˆ’4=0 Γ—2π‘₯^2 (π‘₯βˆ’2)(π‘₯+2)=0 So, π‘₯=2 & π‘₯=βˆ’2 Since π‘₯>0 is given, we consider only π‘₯=2 Finding g’’(𝒙) g’(π‘₯)=1/2βˆ’2/π‘₯^2 g’’(π‘₯)=𝑑/𝑑π‘₯ (1/2βˆ’2/π‘₯^2 ) = 0 – 2 . (βˆ’2) π‘₯^(βˆ’2βˆ’1) = 4π‘₯^(βˆ’3) = 4/π‘₯^3 Putting value π‘₯=2 in g’’(π‘₯) g’’(π‘₯)=4/(2)^3 = 4/8= 1/2 Since g’’ (x) > 0, for x = 2 g(x) is minimum at x = 2 Minimum value of g(π‘₯) is g(π‘₯)=π‘₯/2+2/π‘₯ Putting π‘₯=2 g(2)= 2/2+2/2 = 1 + 1 = 2

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