If the function f:(0,infinite)-->R, f(x)=lnx/x proove that e^pi>pi^e
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You would find very helpful if you would know about the extrema of the function and its monotony.
You should solve the equation f'(x)=0 to find the maximum or minimum value of the function such that:
`f'(x) = (1/x*x - lnx)/(x^2) `
`f'(x) = (1-ln x)/(x^2)`
You need to solve for x the equation f'(x) = 0 such that:
`1-ln x = 0 =gt ln x = 1 =gt x = e`
You need to evaluate f(e) such that:
`f(e) = (ln e)/e =gt f(e) = 1/e`
Notice that the derivative is positive for `x in (1,e)` and it is negative for `x in (e,oo), ` hence the function reaches its maximum at f(e).
Since the function decreases over `(e,oo), ` hence for `eltpi =gt f(e)gtf(pi).`
You shoud substitute `e` and `pi` for x in equation of the function such that:
`1/e gt (ln pi)/pi =gt pi gt e*ln pi`
You may write the inequality in the following way:
`pi*1 gt e*ln pi`
Substituting ln e for 1 yields:
`pi*ln e gt e*ln pi`
Using the properties of logarithms yields:
`ln e^pi gt ln pi^e`
Since the base of logarithm is larger than 1, then `e^pi gt pi^e` .
Hence, using monotony of the given function (the function decreases over `(e,oo))` yields that `e^pi gt pi^e` .
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