keystone problem… keystone problem… next set 18 part 2 © 2007 herbert i. gross

Keystone Problem…Keystone Problem…

next

Set 18 Part 2

© 2007 Herbert I. Gross

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

next


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.

next


next

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


next

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.

next


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.

next

Do not confuse |x| ≤ 1 with x ≤ 1.

next


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.

next

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.

next

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.

next


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.

next

next


next

x = -1

In terms of a picture...

x = 1

y = -1

y = 1

nextnextnextnext

SetS

Cautionnext


For example, ( 1/2 , 3/4 ) is inside the square but | 1/2 | + | 3/4 | > 1.

next

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.

Solution for Problem #1:

The relationship between a number and its absolute value depends on the sign of the number.


nextnext

Hence, there are four separate cases we have to consider…



next

Case 1: (x,y) є S, and: 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1

nextnext

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]

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.

next


In that region, the graph of f1 is given by the equation…

y = f1(x) = 1 – x; where dom f1=[0,1].

next

This is the line segment whose end points are the points (1,0) and (0,1).

next


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

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.

next


Note



next

Case 2: (x,y) є S, and: -1 ≤ x ≤ 0 and 0 ≤ y ≤ 1

nextnext

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]

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.

next



y = f2(x) = x + 1; where dom f2=[0,1].

next

This is the line segment whose end points are the points (-1,0) and (0,1).

next


In terms of a

picture...

(0,1)(-1,0)

nexty = f2(x) = 1 + x, dom f2 = [-1,0]

next

(0,1)(-1,0)

y = f2(x) = 1 + x dom f2 = [-1,0].



next

Case 3: (x,y) є S, and: -1 ≤ x ≤ 0 and -1 ≤ y ≤ 0.

nextnext

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]

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.

next


This is the line segment whose end points are (0,-1) and (-1,0).

next

In that region the graph of f3 is given by the equation…

y = f3(x) = -1 + -x, dom f3 = [-1,0].

next

next


In terms of a

picture...

(0,-1)(-1, 0)

next

y = f3(x) = -1 + -x, dom f3 = [-1,-1]

(0,-1)(-1, 0)

y = f3(x) = -1 + -x, dom f3 = [-1,-1]

next



next

Case 4: (x,y) є S, and: 0 ≤ x ≤ 1 and -1 ≤ y ≤ 0.

nextnext

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]

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.

next


This is the line segment whose end points are (0,-1) and (1,0).

next


y = f4(x) = x – 1; and dom f4 = [0,1].

next

next


In terms of a

picture...

(1,0)

(0,-1)

next

y = f4(x) = x – 1; dom f4 = [0,1] y = f4(x) = x – 1; dom f4 = [0,1]

(1,0)

(0,-1)

next

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.

next


We have enlarged the scale to make the graph easier to view.

next

next


next

(1,0)

(0,1)

(-1,0)

(0,-1)

Juxtaposing the fourgraphs.

next

As shown on the next slide, this

graph does not

represent a function.

next


next

(1,0)

(0,1)

(-1,0)

(0,-1)

For example, the line x = 1/2

intersects the square at 2 points

next

While the graph does not represent a function, it is the union of four graphs

each of which does represent a function.

next


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.

next

next


next

(1,0)

(0,1)

(-1,0)

(0,-1)

next

(1/2,1/2)

(1/2,-1/2)


nextnextnext

f1(x), x є [0,1]

nextnextnext

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) =


nextnextnext

f1(x), x є [0,1]

nextnextnext

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]

keystone problem… keystone problem… next set 18 part 2 © 2007 herbert i. gross

Documents