epdf.pub_the-nuts-and-bolts-of-proofs-third-edition-an-intr

Solutions for the Exercises at the End of the Sections and the Review Exercises 153
25-
- , — , — , — P - - , — I )• I — I — I — r -
9 11 13 15 17 19
c. Consider the function h(x) = lnx/x for x>l. This function is never
negative because Inx > 0 for x > 1. Moreover, h{l) = 0. We can either
graph the function or use the first derivative test to check whether
the function is increasing or decreasing and to find its critical
point(s).
x/x — In X
hXx) = 1
— In X
The function h has a critical point for the value of x such that
1 — In X = 0. Because In x = 1 when x = e, this is the only critical
value for the function h. We will use the second derivative test to
decide what kind of critical point this is.
Therefore,
(-l/x)x2 - 2x(l - Inx) 21nx - 3
h\x) =
h\e) = line-3 2-3 ^3-
This is a negative number. So the value x = e corresponds to a local
maximum of the function. Thus, h{x) < h{e) for all x > 1; that is,
h{x)<h{e) = {ln e)/e^036.... Therefore, /z(x) < 1 for all x>l.
Because (In x)/x < 1 for all x > 1, we can conclude that Inx < x
for all X > 1. In particular. Inn<n for all positive integers, as they are
just real numbers larger than or equal to 1.