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Ex 6.5,3 - Chapter 6 Class 12 Application of Derivatives - Part 24

Ex 6.5,3 - Chapter 6 Class 12 Application of Derivatives - Part 25
Ex 6.5,3 - Chapter 6 Class 12 Application of Derivatives - Part 26
Ex 6.5,3 - Chapter 6 Class 12 Application of Derivatives - Part 27

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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: (vii) g (π‘₯) = 1/(π‘₯^2 + 2)Finding g’(𝒙) g’(π‘₯)=𝑑/𝑑π‘₯ (1/(π‘₯^2 + 2)) g’(π‘₯)=(𝑑(π‘₯^2 + 2)^(βˆ’1))/𝑑π‘₯ g’(π‘₯)=βˆ’1(π‘₯^2+2)^(βˆ’1βˆ’1) Γ— (2π‘₯+0) g’(π‘₯)=βˆ’2π‘₯(π‘₯^2+2)^(βˆ’2) gβ€²(π‘₯)=( βˆ’2π‘₯ )/(π‘₯^2 + 2)^2 Putting g’(𝒙)=𝟎 ( βˆ’2π‘₯ )/(π‘₯^2+2)^2 =0 –2π‘₯=0 Γ—(π‘₯^2+2)^2 –2π‘₯=0 π‘₯=0 Finding g’’(𝒙) g’(π‘₯)=(βˆ’2π‘₯)/(π‘₯^2 + 2)^2 g’’(π‘₯)=(𝑑(βˆ’2π‘₯)/𝑑π‘₯ . γ€– (π‘₯^2 + 2)γ€—^2 βˆ’ (𝑑(π‘₯^2 + 2)^2)/𝑑π‘₯ . (βˆ’2π‘₯))/((π‘₯^2 + 2)^2 )^2 Using quotient rule as (𝑒/𝑣)^β€²=(𝑒^β€² 𝑣 βˆ’ 𝑣^β€² 𝑒)/𝑣^2 =(βˆ’2 (π‘₯^2 + 2)^2βˆ’2 (π‘₯^2 + 2)^(2βˆ’1).𝑑(π‘₯^2 + 2)/𝑑π‘₯ . (βˆ’2π‘₯))/((π‘₯^2 + 2)^2 )^2 =(βˆ’2 (π‘₯^2 + 2)^2βˆ’2 (π‘₯^2 + 2)(2π‘₯ + 0) (βˆ’2π‘₯))/(π‘₯^2 + 2)^4 =(βˆ’2 (π‘₯^2 + 2)^2βˆ’2 (π‘₯^2 + 2)(2π‘₯) (βˆ’2π‘₯))/(π‘₯^(2 )+ 2)^4 =(βˆ’2 (π‘₯^2 + 2)^2+ 8π‘₯^2 (π‘₯^2 + 2))/(π‘₯^(2 )+ 2)^4 =(βˆ’2 (π‘₯^2 + 2)[(π‘₯^(2 )+ 2) βˆ’ 4π‘₯^2 ])/(π‘₯^2 + 2)^4 =(βˆ’2 (π‘₯^2 + 2)(βˆ’3π‘₯^2 + 2))/(π‘₯^2 + 2)^4 =(βˆ’2(βˆ’3π‘₯^2 + 2))/(π‘₯^(2 )+ 2)^3 Putting x = 0 in g’’(x) g’’(0)=(βˆ’2(βˆ’3(0) + 2))/(0^2 + 2)^3 =(βˆ’2(0 + 2))/(2)^3 =(βˆ’4)/8=(βˆ’1)/2 Hence g’’(π‘₯)<0 when π‘₯ = 0 ∴ π‘₯ = 0 is point of local maxima Thus, g(π‘₯) is maximum at 𝒙 = 0 Maximum value of g(𝒙) at x = 0 g(π‘₯)=1/(π‘₯^(2 )+ 2) g(0)=1/(0^2 + 2) = 1/2 Maximum value is 𝟏/𝟐

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.