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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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.