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Keystone Problem…Keystone Problem…
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Set 18 Part 2
© 2007 Herbert I. Gross
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You will soon be assigned five problems to test whether you have internalized the
material in Lesson 18, Part 2 of our algebra course. The Keystone Illustration below is
a prototype of the problems you’ll be doing. Work out the problem on your own.
Afterwards, study the detailed solutions we’ve provided. Notice that several
different ways are presented that could be used to solve each problem.
Instructions for the Keystone Problems
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© 2007 Herbert I. Gross
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As a teacher/trainer, it is important for you to understand and be able to respond
in different ways to the different ways individual students learn. The more ways
you are ready to explain a problem, the better the chances are that the students
will come to understand.
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© 2007 Herbert I. Gross
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Problem #1
Express y as a function of x if
|x| + |y| = 1
Keystone Problem for Lesson 18Keystone Problem for Lesson 18 Part 2Part 2
© 2007 Herbert I. Gross
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The fact that the absolute value of a number cannot be negative tells us that if |
x| + |y| = 1 then both |x| ≤ 1 and |y| ≤ 1.
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© 2007 Herbert I. Gross
Preamble to Our Solution
Namely, if |x| > 1, then the fact that |y| ≥ 0 would mean that |x| + |y| >1. This
would contradict the given information that|x| + |y| = 1. Therefore, we know that |x| ≤ 1.
A similar argument applies to the assumption that |y| > 1.
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Do not confuse |x| ≤ 1 with x ≤ 1.
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© 2007 Herbert I. Gross
Caution
Geometrically speaking, |x| is the distance between x and 0. So to be on the x-axis but no more than 1 unit from 0, x can be
no further than 1 unit to the right of 0 and no further than 1 unit to the left of 0.
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Hence, x ≤ 1 and x ≥ -1 (which we usually write as -1 ≤ x ≤ 1). In a similar way |y| ≤ 1
means that -1 ≤ y ≤ 1.
For example, if x = -3, then the |x| = 3. Hence, x < 1, but |x| > 1.
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In set notation, we want to describe the set S where S = {(x,y): |x| + |y| = 1}.
In terms of the graph of the function, this means we want to describe the set of all
points (x,y) in the xy-plane that satisfy the equation |x| + |y| = 1.
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© 2007 Herbert I. Gross
Geometric Interpretation
Since |x| ≤ 1 and |y| ≤ 1, we see that the set S is contained within the square that is bounded on the left by the line x = -1, on the
right by the line x = 1, below by the line y = -1, and above by the line y = 1.
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© 2007 Herbert I. Gross
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x = -1
In terms of a picture...
x = 1
y = -1
y = 1
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SetS
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Cautionnext
© 2007 Herbert I. Gross
For example, ( 1/2 , 3/4 ) is inside the square but | 1/2 | + | 3/4 | > 1.
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What we have shown is that for any point (x,y) that lies outside the above square, |x| + |y| > 1. However, not every point (x,y) that is in or on the above square satisfies the equation |x| + |y| = 1.
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Solution for Problem #1:
The relationship between a number and its absolute value depends on the sign of the number.
© 2007 Herbert I. Gross
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Hence, there are four separate cases we have to consider…
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Solution for Problem #1:
© 2007 Herbert I. Gross
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Case 1: (x,y) є S, and: 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1
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In this case, |x| = x and |y| = y. Hence, the equation |x| + |y| = 1 becomes x + y = 1;
or, equivalently, y = 1 – x. Stated in the language of functions…
y = f1(x)… where f1(x) = 1 – x, dom f1 = [0,1]
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Geometrically, the set of points (x,y) є S for which |x| + |y| = 1, 0 ≤ x ≤ 1 and
0 ≤ y ≤ 1 is the region which is shown on the following slide.
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© 2007 Herbert I. Gross
In that region, the graph of f1 is given by the equation…
y = f1(x) = 1 – x; where dom f1=[0,1].
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This is the line segment whose end points are the points (1,0) and (0,1).
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© 2007 Herbert I. Gross
In terms of a
picture...(0,1)
(1,0)
nexty = f1(x) = 1 – x; dom f1 = [0,1]
(0,1)
(1,0)
y = f1(x) = 1 – x; dom f1 = [0,1]next
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We do not have to consider any point (x,y) on the line x + y = 1
that lies outside the square because we have already shown
that for all such points, |x| + |y| > 1.
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© 2007 Herbert I. Gross
Note
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Solution for Problem #1:
© 2007 Herbert I. Gross
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Case 2: (x,y) є S, and: -1 ≤ x ≤ 0 and 0 ≤ y ≤ 1
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In this case, |x| = -x and |y| = y. Hence, the equation |x| + |y| = 1 becomes -x + y = 1;
or, equivalently, y = x + 1. Stated in the language of functions…
y = f2(x) where f2(x) = x + 1, dom f2 = [-1,0]
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Geometrically, the set of points (x,y) є S for which |x| + |y| = 1, -1 ≤ x ≤ 0 and
0 ≤ y ≤ 1 is the region which is shown on the following slide.
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© 2007 Herbert I. Gross
In that region, the graph of f2 is given by the equation…
y = f2(x) = x + 1; where dom f2=[0,1].
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This is the line segment whose end points are the points (-1,0) and (0,1).
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© 2007 Herbert I. Gross
In terms of a
picture...
(0,1)(-1,0)
nexty = f2(x) = 1 + x, dom f2 = [-1,0]
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(0,1)(-1,0)
y = f2(x) = 1 + x dom f2 = [-1,0].
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Solution for Problem #1:
© 2007 Herbert I. Gross
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Case 3: (x,y) є S, and: -1 ≤ x ≤ 0 and -1 ≤ y ≤ 0.
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In this case, |x| = -x and |y| = -y. So, the equation |x| + |y| = 1 becomes -x + -y = 1. Multiplying both sides of this equation by -1, we obtain x + y = -1 (or y = -x + -1).Stated in the language of functions…
y = f3(x) where f3(x) = -x + -1, dom f3 = [-1,0]
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Geometrically, the set of points (x,y) є S for which |x| + |y| = 1, -1 ≤ x ≤ 0,
and -1 ≤ y ≤ 0 is the region which is shown on the following slide.
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© 2007 Herbert I. Gross
This is the line segment whose end points are (0,-1) and (-1,0).
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In that region the graph of f3 is given by the equation…
y = f3(x) = -1 + -x, dom f3 = [-1,0].
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© 2007 Herbert I. Gross
In terms of a
picture...
(0,-1)(-1, 0)
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y = f3(x) = -1 + -x, dom f3 = [-1,-1]
(0,-1)(-1, 0)
y = f3(x) = -1 + -x, dom f3 = [-1,-1]
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Solution for Problem #1:
© 2007 Herbert I. Gross
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Case 4: (x,y) є S, and: 0 ≤ x ≤ 1 and -1 ≤ y ≤ 0.
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In this case, |x| = x and |y| = -y. Hence, the equation |x| + |y| = 1, becomes x + -y = 1;
or, equivalently, x = 1 + y or y = x – 1. Stated in the language of functions…
y = f4(x) where f4(x) = x – 1, dom f4 = [0,1]
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Geometrically, the set of points (x,y) є S for which |x| + |y| = 1, 0 ≤ x ≤ 1 and -1 ≤ y ≤ 0 is the region which is shown on
the following slide.
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© 2007 Herbert I. Gross
This is the line segment whose end points are (0,-1) and (1,0).
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In that region, the graph of f4 is given by the equation…
y = f4(x) = x – 1; and dom f4 = [0,1].
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© 2007 Herbert I. Gross
In terms of a
picture...
(1,0)
(0,-1)
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y = f4(x) = x – 1; dom f4 = [0,1] y = f4(x) = x – 1; dom f4 = [0,1]
(1,0)
(0,-1)
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If we now juxtaposition our four graphs, we see that the graph of |x| + |y| = 1 is the
square that is shown on the next slide.
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© 2007 Herbert I. Gross
We have enlarged the scale to make the graph easier to view.
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© 2007 Herbert I. Gross
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(1,0)
(0,1)
(-1,0)
(0,-1)
Juxtaposing the fourgraphs.
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As shown on the next slide, this
graph does not
represent a function.
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© 2007 Herbert I. Gross
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(1,0)
(0,1)
(-1,0)
(0,-1)
For example, the line x = 1/2
intersects the square at 2 points
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While the graph does not represent a function, it is the union of four graphs
each of which does represent a function.
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© 2007 Herbert I. Gross
For example, as shown in the next slide, the line x = 1/2 intersects the square
twice but only once on the red line segment and once
on the blue line segment.
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© 2007 Herbert I. Gross
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(1,0)
(0,1)
(-1,0)
(0,-1)
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(1/2,1/2)
(1/2,-1/2)
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© 2007 Herbert I. Gross
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f1(x), x є [0,1]
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To summarize our results in the language of functions, our graph can be
represented in the form…
f2(x), x є [-1,0]
f3(x), x є [-1,0]
f4(x), x є [0,1]
f(x) =
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© 2007 Herbert I. Gross
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f1(x), x є [0,1]
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And if we now recall the definitions of f1(x), f2(x), f3(x), and f4(x), we may
summarize our results in the form…
f2(x), x є [0,1]
f3(x), x є [-1,0]
f4(x), x є [-1,0]
y =
1 – x, x є [0,1]
x + 1, x є [-1,0]
-x + -1, x є [-1,0]
x – 1, x є [0,1]