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Question

If f(x)=ex24x+3 on [5,5] then log (greatest value of f(x)) is

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Solution

f(x)=ex24x+3f(x)=(ex24x+3).(2x4)f′′(x)=(ex24x+3).2f(x)is0atx=2andf′′(x)isalwayspositive.soatx=2f(x)isminimum.andx=[5,5]thatmeansthatf(x)isdecreasingfrom[5,2]andafterthatitisincreasingtillx=5.sof(5)=e48andf(5)=e8sof(x)wouldbemax.atx=5andlog(greatestvalueoff(x))=log(e48)=48.

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