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NAME DATE BAND

A BSTRACT F UNCTIONS #1 CALCULUS | PACKER COLLEGIATE INSTITUTE

Introduction : In calculus, we are going to spend a lot of time thinking about weird functions what we aregoing to call abstract functions . They arent usually weird looking (well, sometimes they are!), however theyare functions which are based on other functions in strange ways. You have already seen simple versions of this.

original function new function

As you can see, the new function can be created from the original one Each point on the left hand graph isreflected over the x-axis, stretched vertically by a factor of 2, shifted left 4, and moved up 1 and then thenew point is plotted on the right hand graph. The key here is that we created a brand new function based in

some way on the original function . This is something well be doing a lot in this class. Lets start with a simple exercise to get you thinking in this way. Use the left hand graph (LHG) to draw theright hand graph (RHG). If the LHG has a point with a height of 2 or less, plot a point with the same x-value onthe RHG, but give it a height of -3. Similarly, if LHG has a height greater than 2, plot a point on the RHG with aheight of 4. For example, the point (3,1) on the LHG. Thus, we will plot the point (3,-3) on the RHG. Go to it!Plot away!

x

y

x

y

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Below is a graph of 2 y . Were going to create a new graph which calculates the total area between the

graph and the x-axis starting at 0 x . I know this is totes unclear. Look at the bottom of the page to clearthings up.

old function (LHG)

LHGinput

LHGoutput

012

3456

RHGinput

RHGoutput

012

3456

new function (RHG)

Clearing things up! So what are you supposed to do with all this? Youre going to make a new graph based onthe old graph. How? By calculating area Here are s ome examples

x

y

x

y

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This is the old graph. We are going to plotthe point (1,2) on the new graph becauseat 1 x , the total area to the x-axis is 2.

This is the old graph. We are going to plotthe point (2,4) on the new graph becauseat 2 x , the total area to the x-axis is 4.

This is the old graph. We are going to plotthe point (3,6) on the new graph becauseat 3 x , the total area to the x-axis is 6.

Check yourself!On the new graph, with an input of 6, what would the output be? _______On the new graph, with an ouput of 10.6, what would the input have to be? _______

Now that you know what were doing, lets see you work it! Heres a slightly more challenging function on the

LHG. I want you to draw the RHG using the area function. Before you do that, answer couple of questions and fill in the table of values for the RHG

a) Lightly shade in the appropriate area on the LGH for an input of 2.b) On the new graph, with an input of 2, what would the output be? _______ * c) On the new graph, can you guesstimate what the input has to be in order for it to have an output of

about 20? _______

Table of Valuesfor new graph

RHGinput

RHGoutput

0

1

2

3

4

5

6

* To check yourself, your answer should be the solution to 2 3 5 x . Didya get it? Didya?

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LHG RHG: the area function of the LHG(pay attention to the scales on the axes)

Room for scratch work:

Now see if you can answer these questions:On the new graph, with an input of 21.2, what would the output be? _______On the new graph, what would the input have to be exactly to get an output of 20? _______What assumption are you making to answer the previous questions?

One final things before you get going on your own! What if the function goes below the x-axis? Nothing! Youdo everything the same! LUCKY! So lets say the original graph is below, and we want to calculate the new

graph using our abstract area function w hat would the point on the new graph at 3 x ?

Graph 1: original graph Graph 2: using the original graphto calculate the area at 3 x

At 3 x , the shaded area is thesum of two triangles

(2)(2)1 12 2

(1)(1) 2.5 Area

Thus wed plot the point (2,2.5) on

the new graph!

Problems!

1. Original Graph Area Graph

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Take literally no more than 30 seconds to make a prediction of what this Area Graph will look like:

Now open the Geogebra File titled Abstract Functions A2 and drag the point titled Drag Me Copy theresult below:

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Articulate in words why the Area function looks that way from [0,3] and from [3,6]:

2. Original Graph Area Graph

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3. Original Graph Area Graph

Final Test!

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Create a new graph based on this original graph using that whole area thing weve been doing

Original Graph

What is the output of the new function (the area function) going to be at the end when 25 x ? Work space:

Table for New Graph (the Area Graph)

new fxninput

new fxnoutput

0

1

2

3

4

5

new fxninput

new fxnoutput

6

7

8

9

10

new fxninput

new fxnoutput

11

12

13

14

15

new fxninput

new fxnoutput

16

17

18

19

20

new fxninput

new fxnoutput

21

22

23

24

25

Take your answer and multiply it by 3. If you got the right area, that new number should be a perfect cube ! Like 8 or 27. But its not8 or 27. .. I suppose if you took your original answer for the area and divided it by 2, you would get a perfect square . Go figure.

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Draw the new graph (the area graph) below. Note the horizontal scale is 1 but the vertical scale is 5 !

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