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Where is f Continuous?
Date: 10/07/97 at 17:03:59
From: Kelly Stone
Subject: High School Calculus
I have a problem in my calculus class. I really need help and I will
appreciate any help you can give. Here's the problem:
a. Graph the following.
8-|..............
f(x)= {2^x, if x is rational 7-| .
{8 , if x is rational 6-| .
5-| .
4-| .
3-| .
2-| .
1-|.________________
1 2 3 4
b. Besides at x=3, where is is f continuous?
That is the question. I can't get f. Please help!
Date: 10/11/97 at 14:59:56
From: Doctor Chita
Subject: Re: High School Calculus
Hi Kelly:
Have you mistyped your problem? The function as written is supposed to
be a piece-wise function. However, the domain for both pieces is the
same - the rational numbers. Usually, a piece-wise function is made up
of two or more different domains.
I am guessing that the domain of one of the two pieces of your
function is the set of irrational numbers. For discussion, let's
assume, then, that
f(x) = {2^x, if x is rational
{8, if x is irrational
This means that the domain of f is the set of real numbers, since the
union of the rational and irrational numbers is the set of real
numbers.
Consider the two numbers, 1,414, a rational number, and sqrt(2) =
1.4142... , an irrational number. Both numbers correspond to points on
the number line, very close together.
According to the new definition of f, then f(1.414) = 2^(1.414) =
(approximately) 2.665, and f(sqrt(2)) = 8. Consequently, the graph
of f "jumps" between the exponential curve to the horizontal line at
these two points.
This scenario will continue as you move along the x-axis, since the
rational and irrational numbers coexist there. Consequently, f is not
continuous for any interval in its domain.
Unlike other types of discontinuous functions, such as g(x) = 1/x that
has a hole at x = 0, you can't draw a picture of the graph of f
because there are an infinite number of points jumping between the
graphs of 2^x and 8. The only way to represent this function is to use
symbols (the two piece-wise equations) or words.
I hope this answers your question, Kelly. If not, please write back
and we'll try again.
-Doctor Chita, The Math Forum
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