Ex 6.3, 5 (iii) - Chapter 6 Class 12 Application of Derivatives
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Ex 6.3, 5 Find the absolute maximum value and the absolute minimum value of the following functions in the given intervals: (iii) f(π₯) = 4π₯ β 1/2 π₯2 , π₯ β [β2, 9/2] f(π₯) = 4π₯ β 1/2 π₯2
Finding fβ(π)
fβ(π₯)=π(4π₯ β 1/2 π₯^2 )/ππ₯
= 4 β 1/2 Γ2π₯
= 4 β π₯
Putting fβ(π)=π
4 β π₯=0
π₯=4
β΄ π₯=4 is only critical point
Since given interval π₯ β [β2 , 9/2]
Hence , calculating f(π₯) at π₯ = β 2 , 4 , 9/2
Absolute Maximum value of f(x) is 8 at π = 4
& Absolute Minimum value of f(x) is β10 at π = β2
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.
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