Even Answers
Chapter 1 Review
2.
a)
b)
c)
g (2)  3
No horizontal line intersects the graph more than once.
g 1 (2) is the value of x for which g ( x)  2 ; so find where the graph of g
crosses the line y  2 . This happens at x  14 . Thus, g 1 (2)  14
d)
e)
The domain of g 1 is the range of g, which is [-1,3.5].
Flip the graph of g over the line y  x . I don’t have the capcbility in my
office to draw the graph on this web site.
4.
Again I can’t draw graphs of this type. But I would expect the yield to increase as
more fertilizer is used up to a point when too much fertilizer will burn the cxrop
and reduce yield.
6.
Domain is (, 1)  (1, ) . Range is (, 0)  (0, ) Graph the function to
confirm this.
8.
Since you can’t substitute negative numbers or 0 into logs, ln x  0 ; since ln is
an increasing function and since ln1  0 , x  1 . Thus domain is (1,  )
The range is (, )
10.
a)
b)
c)
d)
e)
f)
12.
Shift given graph 8 units to the right.
Flip given graph over x-axis.
Flip over the x-axis and move up 2 units.
Squeeze graph vertically so that every point is half as far from the x-axis
as originally then move the graph down one unit.
Flip the original graph over the line y  x
Take the graph from part e and move it 3 units to the left.
Take graph of y  ln x , shift it 2 units to the right and stretch vertically by a
factor of 3.
14.
Take graph of y  x , flip it over x-axis, and slide it up 2 units.
16.
This is a graph in two pieces. For negative x’s, you have part of the line y   x
and for positive x’s, you take the graph of y  e x and move it down 1 unit.
Pay attention to the scale
18.
2 x  2 if
f ( x)  
2
 1  x if
2  x  1
1  x  1
20.
( f g h)( x)  F ( x) 
1
x x
where f ( x) 
1
, g ( x)  x 2  x , and
x
h( x )  x
22.
24.
a)
c  6n  3000
b)
c)
The slope is 6 and it represents the cost to produce one toaster oven.
The y-intercept is 3000 and it represents the overhead costs.
We want to solve the given equation for x:
x 1
y
2x 1
y (2 x  1)  x  1
2 xy  y  x  1
2 xy  x  1  y
x(2 y  1)  1  y
1 y
x
2 y 1
Using the usual function notation, f 1 ( x) 
26.
a)
b)
c)
1 x
2x 1
e x  5 , so x  ln 5
ln x  2 , so x  e2
x
ee  2 , so e x  ln 2 , which implies that x  ln(ln 2)
tan 1 x  1 , so x  tan1
d)
All of these problems are essentially the definitions of the functions involved
28.
a)
b)
c)
30.
It takes about 4 years to reach 900
We want to solve the given equation for t:
100, 000
P
100  900e  t
P(100  900et )  100,000
100, 000
100  900et 
P
100, 000
100, 000  100 P
900e t 
 100 
P
P
100, 000  100 P 1000  P
et 

900 P
9P
1000  P
t  ln
9P
1000  P
t   ln
9P
This formula gives the n time it takes for the population to reach size P.
To find how long it takes for the population to reach 900, substitute 900
for P in this last equation:
t  4.39 years
From slowest growth to fastest the functions are log a x , x a , and a x .
ln x
To graph y  log a x , you have to graph y 
ln a