1873_solutions

# Defined on an interval s and that f x 0 for every x s

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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 atsb 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.

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