Exercise 1.7.19 solution:
Exercise 1.7.18 solution:
Exercise 1.7.17 solution:
Exercise 1.7.15 solution:
Exercise 1.7.13 solution:
Exercise 1.7.12 solution:
Exercise 1.7.11 solution:
Exercise 1.7.10 solution:
Exercise 1.7.9 solution:
Exercise 1.7.8 solution:
Exercise 1.7.7 solution:
Exercise 1.7.6 solution:
Exercise 1.7.5 solution:
Since 
, f is an odd
function.
Exercise 1.7.4 solution:
So the graph must be symmetric with respect to the origin.
Exercise 1.7.3 solution:
So the graph must be symmetric with respect to the x-axis.
Exercise 1.7.2 solution:
Therefore, the function is even.
Exercise 1.7.1 Solution:
Symmetric with respect to the y axis.
Exercise 1.6.5 solution:
for 
for 
.
Exercise 1.6.4 solution:
for 
Interchanging x and y yields 
for 
.
Solving for y yields 
. Since 
, we know it must be 
and 
.
So 
for 
.
Exercise 1.6.3 solution:
Exercise 1.6.2 solution:
Thus, 
Exercise 1.5.5 solution:
for 
and
. In interval
notation, the domain of 
is
.
Exercise 1.5.4 solution:
for 
and 
. Thus 0≤x≤16. In interval
notation, the domain of 
is [0,16].
Exercise 1.5.3 solution:
for 
.
Exercise 1.5.2 solution:
Exercise 1.5.1 solution:
Exercise 1.4.5 solution:
Domain 
, Range 
.
Exercise 1.4.4 solution:
The implicit domain is 
. To find the range,
replace 
with y and
solve for x to get
. So y
cannot equal to 2. Thus, the range of
the function is 
.
Exercise 1.4.3 solution:
Since we cannot divide by 0, we know that x cannot
equal to either 0 or 1. And since we cannot
take the square root of a negative number (remember, output numbers must be
real), then 
. Taken together,
these requirements mean that 
. Thus the implicit
domain is 
.
Exercise 1.4.2 solution:
Since the quantity under the radical must be non-negative, 
.
Thus the implicit domain is 
.
Exercise 1.4.1 solution:
x can take any value except 3, thus the domain is, in interval notation,
Exercise 1.3.2 solution:
The graph of 
The graph of 
Exercise 1.3.1 solution:
For the equation 
, there is only one possible value for y given any
particular value of x, namely, the square of x. So y is a function of x.
But for the equation 
, there can be two values of y for some values of x. For example, for x = 4, y
could have either the value of +2 or -2. Thus, for the second equation, y is not
a function of x.
Exercise 1.2.3 solution:
We want to find 
given that 
.
We know that 
. Then placing 
into each of the
parentheses gives:
And we already know that 
.
Thus 
Exercise 1.2.2 solution:
