Example 39 - Chapter 6 Class 12 Application of Derivatives (Term 1)
Last updated at April 19, 2021 by Teachoo
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Example 39 You are here
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Transcript
Example 39 Find the absolute maximum and minimum values of a function f given by π (π₯) = 2π₯3 β 15π₯2 + 36π₯ +1 on the interval [1, 5]. f(π₯) = 2π₯3 β 15π₯2 + 36π₯ + 1 Finding fβ(π) fβ(π₯)=6π₯^2β30π₯+36 Putting f β(π)=π 6π₯2 β 30π₯ + 36 = 0 6(π₯^2β5π₯+6)=0 (π₯^2β5π₯+6)=0 (π₯^2β2π₯β3π₯+6)=0 π₯(π₯β2)β3(π₯β2)=0 (π₯β2)(π₯β3)=0 Hence. π = 2 or π = 3 Since we are given interval [π , π] Hence, calculating f(π₯) at π₯ = 1, 2, 3, 5 Hence, Absolute maximum value is 56 at π=π Absolute minimum value is 24 at π=π
