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Inequalities

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					CHAPTER 3
Functions, Equations, and
Inequalities
   Section 3.5 - Inequalities




 for values of x in which everything is defined
Example: If -3x < 15 then …
                                 x > -5
If a function f is always either increasing or
decreasing (not both) on its domain,
then it’s one-to-one.
Proof by Contraposition:          This is new!!
If a function f is always either increasing or
decreasing (not both) on its domain,
then it’s one-to-one.
Proof by Contraposition:               This is new!!

What is the contrapositive of the expression p  q ?
                     ~q  ~p
What is the contraposition of the conditional
expression that we’re trying to prove?
                                    De Morgan’s Law
~(the function is one-to-one) 
        ~(either increasing or decreasing on its domain)
If f is not one-to-one, it’s not increasing
and not decreasing on its domain.
If a function f is always either increasing or
decreasing (not both) on its domain,
then it’s one-to-one.
Proof by Contraposition:
(show that if f is not one-to-one, it’s not increasing
and not decreasing).
Assume f is not 1-1.
If a function f is always either increasing or
decreasing (not both) on its domain,
then it’s one-to-one.
Proof by Contraposition:
(show that if f is not one-to-one, it’s not increasing
and not decreasing).
If a function f is always either increasing or
decreasing (not both) on its domain,
then it’s one-to-one.
Proof by Contraposition:
(show that if f is not one-to-one, it’s not increasing
or decreasing).

So if f is not one-to-one, it is not increasing
and not decreasing.
    Therefore, by contraposition:
If f is either increasing or decreasing then it is
one-to-one.
Q.E.D. = “Quod Erat Demonstratum” means “So it is demonstrated”
So, for every function h(x) which is either
increasing or decreasing on its domain…
  •Is one-to-one
  •Passes the horizontal line test
  •Has inverse function h-1(x)
  •f(x) = g(x) if and only if h f(x) = h g(x)
               (This is biconditional)
 Section 3.5 - Inequalities

A function f(x) is one-to-one iff
     it has an inverse f -1(x)


If a function f is always either
increasing or decreasing (not both)
on its domain,
      it has an inverse f -1(x)
Section 3.5 - Inequalities
If a function f is always increasing on
its domain, then f –1 is increasing on
its domain.

If a function f is always decreasing on
its domain, then f –1 is decreasing on
its domain.
    Proof by Contradiction!!
If a function f is always increasing on its
domain, then f –1 is increasing on its domain.
Assume f is always increasing
but f -1 is not always increasing.
Then f -1must be not increasing somewhere.

Proof by Contradiction: Assume the
premise is true and conclusion is false and
reach a contradiction
If a function f is always increasing on its
domain, then f –1 is increasing on its domain.
This contradicts our assumption that f is
increasing on its domain.
Thus, f -1 must be increasing on its domain.
 Section 3.5 - Inequalities

If a h(x) increasing on its domain, then


If a h(x) decreasing on its domain, then
Section 3.5 - Examples



Solve:
Section 3.5 - Homework


   Read Section 3.5 and do 1 – 19
Section 3.6 - Examples

Solve:

1.   “factor” by chunking
2.   Using the Zero-Product Property
 Section 3.6 - Warm Up

The sum of a positive number and
 its reciprocal is less than two.
 What is the number?
 Section 3.4 - Quick Review:

What does it mean to say a function is
 continuous on an interval?
    Give an example of a function which
    is continuous.
    Give an example of a function which
    is not continuous.
What does the Intermediate Value
 Theorem say?

				
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posted:9/20/2011
language:English
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