Estimation quiz!
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07-30-2020, 03:10 PM
Post: #9
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RE: Estimation quiz!
Another way, let f = ln(e^x/x^e) = x - e*ln(x)
f' = 1 - e/x = 0 ⇒ x=e // locate extremum f'' = e/x^2 > 0 // 2nd derivative test, f(e) is minimum if f x ≠ e: f(x) > f(e) = 0 ⇒ e^x > x^e → e^pi > pi^e |
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