University Algebra: Homework 6 - Function, Domain, Range Analysis

This document presents detailed solutions for College Algebra Homework 6. The solutions cover various topics, including describing the graph of a circle and identifying its center and radius, determining the domain and range of relations, and determining if a relation is a function. The document also includes solutions to determine whether equations define y as a function of x. Furthermore, it provides solutions for finding the domain of different functions, including those with square roots and rational expressions. Finally, the document provides solutions for computing and simplifying expressions involving given functions f(x), g(x), and p(x). The solutions are well-explained, step-by-step and provide clear explanations for each problem, making it a valuable resource for students studying college algebra. The assignment covers a range of core concepts within the subject, ensuring a thorough understanding of function-related concepts.

Solution 1: Given equation is . Let’s simplify it and we get
The equation (1) is of the form with center and radius .
Hence, given equation is a circle whose center is and radius . The graph of
circle is
Solution 2a: We know that in an ordered pair,, the domain is the set of the first number
in every pair (those are the x-coordinates) and the range is the set of the second number
of all the pairs (those are the y-coordinates).
Hence, domain of the relation is and range is
.
We know that a function is a relation in which each input has exactly one output. But the
input 4 and 9 has more than one output. Hence, given relation is not a function.
Solution 2b: We know that in an ordered pair, the domain is the set of the first number in
every pair (those are the x-coordinates) and the range is the set of the second number of
all the pairs (those are the y-coordinates).
Hence, domain of the relation is and range is
.
We know that a function is a relation in which each input has exactly one output and in
the above relation each input has exactly one output. Hence, the given relation is a
function.
Solution 3a: Given equation is . Simplify we get
. We can see that for every value of x, there exist unique values of y. Hence,
given equation defines y as a function of x.
Solution 3b: Given equation is . Simply we get
We can see that two values exists for every x. hence equation is not defines y as a
function of x.
Solution 3c: Given equation is . Simplify we get
We can see that we have two possible values to y for each value of x. hence equation is
not defines y as a function of x.
Solution 4a: Given expression is . We know that the domain of a
function is the set of all input values for which function is real and defined.

Now, solve we get . Since, function is not defined at
. Hence, domain of the function is the set of real values except
. In interval notation, the domain of the function is
.
Solution 4b: Given expression is .
Since, function is not defined at and to be real for
Hence, the domain of the function is
Solution 4c: Given expression is .
Solve we get and function to be real for . So, the
domain of the function will be all values except and
Hence, the domain of the function is
Solution 5: Given that
a):
b):
c):
d):