Ex 6.3

Chapter 6 Class 12 Application of Derivatives
Serial order wise

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### Transcript

Ex 6.3, 10 Find the maximum value of 2π₯3 β 24π₯ + 107 in the interval [1, 3]. Find the maximum value of the same function in [β3, β1]. Let f(π₯)=2π₯^3β24π₯+107 Finding fβ(π₯) fβ(π₯)=π(2π₯^3 β 24π₯ + 107)/ππ₯ = 2 Γ 3π₯^2β24 = 6π₯^2β24 = 6 (π₯^2β4" " ) Putting f(π₯)=0 6 (π₯^2β4" " )=0 π₯^2β4" = 0 " π₯^2=4 π₯=Β±β4 π₯=Β±2 Thus, π₯=2 , β 2 Since x is in interval [1 , 3] π₯ = 2 is only Critical point Also, since given the interval π₯= β [1 , 3] We calculate f(x) at π₯= 1 , 2 & 3 Hence maximum value of f(π)=ππ at π = 3 in the interval [1 , 3] For the interval [βπ , βπ] π₯ = β2 is only Critical Point Also, since given the interval π₯= β [β3,β1] We calculate f(x) at π₯= β1 , β2 & β3 Hence maximum value of f(π)=πππ at π = β2 in the interval [β3,β1]

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#### Davneet Singh

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.