Let f be a function that is increasing and concave up, and g be a
function that is increasing and concave down. Graphically, we know that f
and g are as follows.
Since f increases faster than g, we may intuitively think that f will
eventually be greater than g. The main question of this short note is : Must
there exist a point x1 such that f (x1) > g(x1) ?
Consider the functions f(x) = x - ln (x) and g(x) = x + ln (x), where
x > 1. We clearly have f '(x) = 1- , f "(x) = , i.e. f is increasing and
concave up, and g '(x) = 1+ , g "(x) = - , i.e. g is increasing and concave
down. But we always have f(x) g '(x0). Does the point x1 hypothesized above exist? We
will show that now it does.
To simplify our proof, let h = f - g. Then h '(x0) > 0 and h"(x) > 0 for all x.
Consider any x1 > x0 - . By Taylor series, we have
h(x1) = h(x0) + h'(x0)(x1 - x0) + h"(c)(x1 - x0)2 , where c ? x0 , x1)
?h(x0) + h'(x0)(x1 - x0)
> h(x0) + h'(x0)(x0 - - x0 )
= 0, which concludes the proof.
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