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1.3, Problem #2
Explain how the following graphs are obtained from the graph of 
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(a)  |
This function will be 5 times "steeper" than the original, . |
This is a stretching transformation |
(b)  |
This function will be shifted 5 units to the right. | This is a translation (or lateral translation). |
(c)  |
This function will be the mirror image of
with respect to the x-axis. (It will be up-side-down.) |
This is a reflecting transformation |
(d)  |
This function will be 5 times "steeper" and flipped up-side-down. | This is both a lateral translation and a reflecting translation. |
(e)  |
This will compress the function horizontally. | This is a stretching transformation |
(f)  |
This function is made 5 times "steeper" and is translated 3 units down. | This is both a stretching and a vertical transformation. |
1.3, Problem #10.
Graph 
not by
plotting points, but by starting with the graph of one of the standard functions
given in Section 1.2.
Think of the function as:  |
The Addition form will make it more obvious what seperate transformations are taking place. |
 and 
are the two component parts. |
Break the function into its two parts to see what role each part plays |
| is the cosine
function with a reflecting transformation (up-side-down) |
Sketch an up-side-down cosine curve. |
| Sketch the horizontal curve
on the same sketch that your up-side-down cosine curve is on. |
added to any curve
will have the effect of raising the curve 2 units. This is a translation. |
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Combine the two sketches by drawing the "sum curve". Using Maple, you
would use the following commands:
> with(plots); > plot([2,-cos(x),2-cos(x)],x=-6..20); The red curve is the graph of |
1.3, Problem #38.
Given that 
and 
, find
the functions: (a) 
,(b) 
,(c) 
,
and (d) 
and
their domains.
(a) 
means: 
, so we are to substitute the g function wherever the x
appears in the f function. This gives us:  .
This simplifies to:  ,
so  . |
The domain of a composite function is the intersection of
the domains of the functions being composed. Since  ,
and  ,
then the domain of
is the intersection of the two above domains. Therefore the domain of
is  . |
(b) 
means:  ,
so we are to substitute the f function wherever the x appears
in the g function. This gives us:  .
This simplifies to:  ,
so  .
Note that  . |
The domain of a composite function is the intersection of
the domains of the functions being composed. Since ,
and ,
then the domain of
is the intersection of the two above domains. Therefore the domain of
is . |
(c) 
means:  ,
so we are to substitute the f function wherever the x appears
in the f function. This gives us:  .
This simplifies to:  ,
so  . |
Since we are composing this function with itself, its domain
is simply the domain of the function. There fore the domain of
is  . |
(d) 
means:  ,
so we are to substitute the g function wherever the x appears
in the g function. This gives us:  .
This simplifies to:  ,
so  . |
Since we are composing this function with itself, its domain
is simply the domain of the function. There fore the domain of
is all real numbers. |
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