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Unformatted text preview: ity as
fx
f x.
x 0, Ý we shall show that g x Suppose that x 0. We now apply the mean value theorem to the function f on the interval 0, x to choose
a number c between 0 and x such that
fx f0
f c.
x0
Since c x and since the function f is increasing we know that f c
f x and we have therefore
deduced that the inequality
fx
fx
x
holds.
14. Suppose that f is defined on an open interval S and that f x 0 for every number x S. Suppose that
a S. Prove that if x S and x a then
fx fa x af a.
In other words, explain why, to the right of the point a, f a , the graph of f lies below the tangent line to the
graph at a, f a .
Suppose that x S and that a x. Using the mean value theorem we choose a number c between
a and x such that
fx fa
f c.
xa
Now since f is everywhere negative in S the function f must be strictly decreasing and so
fx fa
f c f a
xa
from which we deduce that
fx fa x af a.
15. Suppose that f is defined on an open interval S and that f x 0 for every number x
and b belong to S and that a b. Suppose that
fb fa
gx fx fa
xa
ba
for all x
a, b . S. Suppose that a a. Prove that there exists a number c
a, b such that g c 0.
We apply Rolle’s theorem to g exactly as in the proof of the mean value theorem.
b. Prove that the function g is strictly decreasing on the interval a, b .
The function g is strictly decreasing because g x f x 0 for each x.
c. Prove that the function g is strictly increasing on a, c and strictly decreasing on c, b .
Since g c 0 and g is strictly decreasing we know that g x 0 whenever a
g x 0 whenever c x b. x c and d. Prove that g x 0 for every x
a, b .
Since g a 0 and g is strictly increasing on a, c we have g x 0 whenever a x 244 c. Since g b 0 and g is strictly decreasing on c, b we have g x 0 whenever c
e. Prove that the straight line segment that joins the points a, f a and b, f b
graph of f that lies between the two points a, f a and b, f b .
This exercise is asking us to prove that
fb fa
x a fx
fa
ba
whenever a x b and this fact follows at once from Part d. x b. lies under the part of the 16. A function f defined on an interval S is said to be convex on S if whenever a, x and b belong to S and
a x b we have
fx fa
fb fx
.
xa
bx
Prove that if f is differentiable on an interval S then f is convex on S if and only if the function f is
increasing.
This exercise is asking us for two proofs.
Part 1: We assume that f is a convex function on an interval S and we want to prove that f is
increasing on S. We begin by making the observation that if a and s and b and t belong to S and if
atsb a s t b then
ft fa
fs ft
fb fs
ta
st
bs
Suppose now that a and b are any two numbers in S and that a b. If t and s are numbers unequal
to a and b but sufficiently close to a and b, respectively, then t s and, regardless of the order in
which a and t appear and the order in which s and b appear we have
ft fa
fs fb
.
ta
sb
Therefore
ft fa
fs fb
lim
f b
f a lim
ta
ta
sb
sb
Part 2: We...
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This note was uploaded on 11/26/2012 for the course MATH 2313 taught by Professor Staff during the Fall '08 term at Texas El Paso.
 Fall '08
 STAFF
 Math, Calculus

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