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y=logex/x^2 then minimum value of y is?kindly explain the answer and thanks in advance.

y=logex/x^2 then minimum value of y is?kindly explain the answer and thanks in advance.

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1 Answers

Asad Rehman
27 Points
5 years ago
The given function isy=(1x)xy=(1x)xNow,taking logarithm from both sides,we get..logy=−xlogxlogy=−xlogxDifferentiating both sides w.r.t x….1ydydx=−[1+log]1ydydx=−[1+log]dydx=−(1x)x[1+logx]dydx=−(1x)x[1+logx]For max or min of f(x) ,put…f′(x)=0=−(1x)x[1+logx]f′(x)=0=−(1x)x[1+logx]⇒logx=−1=log1e⇒logx=−1=log1e⇒x=1e⇒x=1eAgain,f′′(x)=−(1x)x1x−(1x)x[1+logx]2f″(x)=−(1x)x1x−(1x)x[1+logx]2Now,f′′(1e)=−[e1e+1+0]f″(1e)=−[e1e+1+0]=−e1e+1>0=−e1e+1>0This shows that f(x) has minimum value at x=1ex=1eHence,the required minimum value of f(x) ise1e

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