Suffolk County Community College
Michael J. Grant Campus
Department of Mathematics
Wednesday, May 4, 2016
MAT 124, CRN 21802
Fundamentals of Precalculus I
Final Exam: Solutions and Answers
Instructor:
Name: Alexander Kasiukov
Office: Health, Science and Education Center, Room 109
Phone: (631) 851-6484
Email: [email protected]
Web Page: http://www2.sunysuffolk.edu/kasiuka/
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Problem 1. Suppose
set A = {cat, dog, rat, cow} and
set B = {Rex, Mickey, Skip, Spike}. Define a function “Name” to have domain A, range B
and graph
{(cat, Skip), (dog, Rex), (rat, Mickey), (cow, Spike)} .
(1). What is Name(dog)?
Space for your solution:
By definition, Name(dog) is the second element of the only ordered pair in the graph of the
function Name that has “dog” as its first element. In our case, such pair is the pair (dog, Rex),
therefore Name(dog) = Rex.
(2). What is the image of the function Name?
Space for your solution:
By definition, the image of a function f is the set of all the elements in its range that have
the form f (x), where x is an element of the domain of f :
Im f = {f (x) : x ∈ Dom f } .
In our case,
Im Name = {Name(cat), Name(dog), Name(rat), Name(cow)} =
{Skip, Rex, Mickey, Spike} .
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(3). Can the function Name be inverted? If yes, find the domain, range and graph of the
inverse. If no, explain why not.
Space for your solution:
A function f can be inverted if and only if for every y ∈ Rg f , there exists unique x ∈ Dom f ,
such that f (x) = y. The “unique” part means that f is one-to-one and the “exists” part
means that f is on-to, in other words Rg f = Im f .
In our case, the function Name satisfies the above condition.
It is on-to because its image (found in the previous problem) coincides with its range:
Im Name = {Skip, Rex, Mickey, Spike} = {Rex, Mickey, Skip, Spike} = B = Rg Name.
By direct examination, it is clear that the function Name is also one-to-one, i.e. for every
y ∈ Rg Name, there is only one element x ∈ Dom Name, such that Name(x) = y.
Dom Name−1 = Rg Name = {Rex, Mickey, Skip, Spike} ,
Rg Name−1 = Dom Name = {cat, dog, rat, cow} ,
Gr Name−1 = {(Skip, cat), (Rex, dog), (Mickey, rat), (Spike, cow)} .
Problem 2. Denote R the set of all real numbers. Define a function with the domain R
and range R by the formula
f (x) = −3x2 + 6x + 15
(1). Describe, using the set builder notation, the graph of the function f .
Space for your solution:
Whenever a function is defined by a formula f (x), the graph of that function is the set
Gr f = x, f (x) : x ∈ Dom f .
In our case, the graph is the set
Gr f =
x, −3x2 + 6x + 15 : x ∈ R .
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(2). Sketch the graph of the function f and find the image of the function f .
Space for your solution:
Complete the square:
−3x2 +6x+15 = −3 x2 −2x−5 = −3 x2 +2x(−1)+(−1)2 −(−1)2 −5 = −3 (x−1)2 −6 .
1. start with x2 ,
2. shift right by 1 to get (x − 1)2 ;
3. shift down by 6 to get (x − 1)2 − 6;
4. stretch vertically by factor of 3 to get 3 (x − 1)2 − 6 ;
5. flip the graph around the Ox-axis to get −3 (x − 1)2 − 6 .
If you trace the vertex of the parabola from its original position at (0, 0), you will end up at
the point (1, 18). Therefore
Im f = (−∞, 18].
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Problem 3. Consider the function with the range R, defined by the formula
f (x) =
x3 − x2 + x − 1
x2 + 2x + 1
for all x ∈ R, for which the above formula makes sense.
(1). What is the domain of the function f ?
Space for your solution:
The above formula makes sense if and only if the denominator of the fraction is not zero.
The denominator is zero if and only if x2 + 2x + 1 = 0 ⇔ (x + 1)2 = 0 ⇔ x = −1.
Therefore the domain of the function f is
Dom f = {x ∈ R : x 6= −1} .
(2). Find all x intercepts of the function f (x).
Space for your solution:
To find the x intercepts of the function f (x), we need to determine the roots of its numerator
x3 −x2 +x−1. Since it is a polynomial of degree 3, we need to use the rational roots method.
Trying out all possible expressions of the form
divisor of − 1
= {1, −1}
divisor of 1
yields the root x = 1.
Long division of the polynomial x3 − x2 + x − 1 by the polynomial x − 1 gives:
2
3 x 2 + 0x + 1
x−1
x −x + x−1
x3 − x2
x−1
x−1
0
Since the resulting ratio x2 + 0x + 1 = x2 + 1 has no real roots, x = 1 is the only real root
of the polynomial x3 − x2 + x − 1. Since x = 1 is in the domain of the function f (x), it is
an x intercept of the function f (x) (and the only one).
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(3). Write f (x) as a sum of a polynomial and a proper rational function.
Space for your solution:
Long division of the polynomial x3 − x2 + x − 1 by the polynomial x2 + 2x + 1 gives:
x−3
3
2
2
x − x + x−1
x + 2x + 1
x3 + 2x2 + x
−3x2 + 0x − 1
−3x2 − 6x − 3
6x + 2
Therefore
x 3 − x2 + x − 1
6x + 2
f (x) =
=x−3+ 2
.
2
x + 2x + 1
x + 2x + 1
(4). Sketch the graph of the function f (x).
Space for your solution:
Using the results of the previous sub problems, we get:
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Problem 4. Solve equation 3 + log2 x = log2 (3 + x).
Space for your solution:

 3 = log2 3+x
x
3+x>0
3+log2 x = log2 (3+x) ⇔ 3 = log2 (3+x)−log2 x ⇔

x>0
3+x
8= x
8x = 3 + x
7x = 3
⇔
⇔
⇔
⇔ x = 73 .
x>0
x>0
x>0

3+x
 23 = 2log2 x
⇔
x > −3

x>0
Problem 5. Using the technique of graph transformations, sketch the graphs of 3 + log2 x
and log2 (3 + x) in the same (x, y)-coordinate system. Then compare your answer with that
of the previous problem.
Space for your solution:
The graph of
3,
and log2 (3
3 + log2 x is
+ x) is the
the result of shifting log2 x
result of shifting log2 x left
up
by
by
3:
These two resulting grapsh do intersect at the point consistent with the previous solution
x = 37 .
Problem 6. Solve the equation 52x =
1
.
3x−1
Space for your solution:
1
52x = 3x−1
⇔ 52x = 3−(x−1) ⇔ 25x = 3−x 31
x
log75 75 = log75 3 ⇔ x = log75 3.
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⇔ 25x 3x = 3
⇔ 75x = 3
⇔
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