Maths - 2000 Paper 1 Q7 (C)(II)

alan4cult

f(x) = (ln x)/x where x > 0

1. Show that the maximum of f(x) occurs at the point (e, 1/e)
2. Hence, show that x^e ≤ e^x for all x > 0 (MY PROBLEM)

How do I show this?

ZorbaTehZ

In part one you showed that the maximum was at (e, 1/e)
Therefore the greatest y value of the graph is 1/e (the y-value of the maximum)
Therefore the function (or f(x), which is actually the y-value of the graph) will always be less than or equal to that maximum, else it isn't actually the maximum.

(Ln x)/x ≤ 1/e
e(ln x) ≤ x
ln x^e ≤ x
(x^e) ≤ e^(x)
x^e ≤ e^x

alan4cult

Thanks

God, I'm so stupid
I'd a prob done this, but that's just my method!
ln x^e ≤ x
ln x^e ≤ xln e
ln x^e ≤ ln e^x
Then Drop the logs
x^e ≤ e^X

ZorbaTehZ

It's the same as mine, since the log rule is:
Log_Base (Answer) = Power

Congruent to:
Answer = Base^Power

So in my case:
Log_e (x^e) ≤ x
x^e ≤ e^x

EDIT:Error corrected.