Definition: Let f be a real-valued function of a real variable and let M be any real number. The function Mf, called the multiple of f by M or M times f, is the real-valued function with the same domain as f that is defined by the rule Mf(x) = M(f(x)) for each x in the domain of f. Assume that f is a real-valued function of a real variable, and prove the following statement. If f is increasing on a set S and if M is any negative real number, then Mf is decreasing on S. function on a set S of real numbers, and suppose M is any negative Proof: Suppose f is any increasing number. [We will show that Mf is decreasing on S.] Let x₁ and x₂ be any real numbers in S such that x₁ < x₂. Since f is increasing on S, f(x₁)< f(x₂). Multiplying both sides by M gives that M(xi) negative X M(x₁) ---Select--- -f(x₁) M(x₁) M(f(x)) M(x₂) X because M is --Select--- ---Select--- -f(x₁) -f(x=) M(x1) M(x₂) M(f(x:)) M(f(x=)) V By definition of Mf, (MF)(x₁) = ---Select--- and (MF) (x₂) = Thus, by substitution, (Mf)(x₁) * (MF) (x₂). Since x₁ and x₂ were arbitrarily chosen real numbers in S such that x₁ < x₂, we can conclude that Mf is decreasing on S. V real
Definition: Let f be a real-valued function of a real variable and let M be any real number. The function Mf, called the multiple of f by M or M times f, is the real-valued function with the same domain as f that is defined by the rule Mf(x) = M(f(x)) for each x in the domain of f. Assume that f is a real-valued function of a real variable, and prove the following statement. If f is increasing on a set S and if M is any negative real number, then Mf is decreasing on S. function on a set S of real numbers, and suppose M is any negative Proof: Suppose f is any increasing number. [We will show that Mf is decreasing on S.] Let x₁ and x₂ be any real numbers in S such that x₁ < x₂. Since f is increasing on S, f(x₁)< f(x₂). Multiplying both sides by M gives that M(xi) negative X M(x₁) ---Select--- -f(x₁) M(x₁) M(f(x)) M(x₂) X because M is --Select--- ---Select--- -f(x₁) -f(x=) M(x1) M(x₂) M(f(x:)) M(f(x=)) V By definition of Mf, (MF)(x₁) = ---Select--- and (MF) (x₂) = Thus, by substitution, (Mf)(x₁) * (MF) (x₂). Since x₁ and x₂ were arbitrarily chosen real numbers in S such that x₁ < x₂, we can conclude that Mf is decreasing on S. V real
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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I struggle the most starting from "Multiplying both sides by M gives that..." and below
Transcribed Image Text:This exercise uses the following definition.
Definition:
Let f be a real-valued function of a real variable and let M be any real
number. The function Mf, called the multiple of f by M or M times f, is
the real-valued function with the same domain as f that is defined by
the rule
Mf(x) = M(f(x)) for each x in the domain of f.
Assume that f is a real-valued function of a real variable, and prove the following statement.
If f is increasing on a set S and if M is any negative real number, then Mf is decreasing on S.
function on a set S of real numbers, and suppose M is any negative
Proof: Suppose f is any increasing
number.
We will show that Mf is decreasing
Let x₁ and x₂ be any real numbers in S such that X₁ < x₂.
Since f is increasing on S, f(x₁)<
f(x₂).
Multiplying both sides by M gives that M(x₁)
negative
on S.
M(x₁)
---Select---
-f(x:)
M(x₁)
M(f(x:))
X M(x₂)
and (Mf)(x₂) = ---Select--- ♥
because M is
---Select---
-f(x₁)
-f(x₂)
M(x1)
M(x₂)
By definition of Mf, (Mf)(x₁) = ---Select---
Thus, by substitution, (Mf)(x₁) < * (MF) (x₂).
Since x₁ and x₂ were arbitrarily chosen real numbers in S such that X₁ < x₂, we can conclude that Mf is decreasing
on S.
M(f(x:))
M(f(x₂))
real
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