![Ex 6.5, 10 - Find max value of 2x3 - 24 x + 107 in [1, 3]](https://d1avenlh0i1xmr.cloudfront.net/31eed117-cb7b-4cc8-8cc9-53255594a1f1/slide30.jpg)
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Ex 6.3,10 - Chapter 6 Class 12 Application of Derivatives
Last updated at May 29, 2023 by Teachoo
Ex 6.3
Ex 6.3, 1 (i)
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Ex 6.3, 1 (ii)
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Ex 6.3,10 You are here
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Ex 6.3, 27 (MCQ)
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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 [β3, β1]. Let f(π₯)=2π₯^3β24π₯+107 Finding fβ(π₯) fβ(π₯)=π(2π₯^3 β 24π₯ + 107)/ππ₯ = 2 Γ 3π₯^2β24 = 6π₯^2β24 = 6 (π₯^2β4" " ) Putting f(π₯)=0 6 (π₯^2β4" " )=0 π₯^2β4" = 0 " π₯^2=4 π₯=Β±β4 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)=ππ at π = 3 in the interval [1 , 3] For the interval [βπ , βπ] π₯ = β2 is only Critical Point Also, since given the interval π₯= β [β3,β1] We calculate f(x) at π₯= β1 , β2 & β3 Hence maximum value of f(π)=πππ at π = β2 in the interval [β3,β1]
