Name: Solutions
Due Date: Friday, December 2nd in Class
Extra Credit 5/Final Practice Exam
Definitions and Concepts
Directions: Complete the following definitions by inserting the word/s or mathematical text as indicated by the
labeled underlined blanks. If asked, formulate your own definition in your own words (as long as it is still correct of
course).
1. A function f(x) is a (a) function if in addition to passing the
function of x, it also passes the (c) line test.
(b)
line test, which is given since f(x) is a
(a) 1-1
(b) vertical
(c) horizontal
2. Given functions f and g we define the domain of the composition (f ◦ g)(x) as:
The set of all elements, x, in the domain of g such that g(x) is in the domain of f.
3. Let F(x) be a polynomial function. If f(c) = 0 or (c, 0) is a point on the graph of f, then
a.
is a factor of f.
We define the
b.
of c as the number of times
c.
appears in the factored form of f.
(a) x − c
(b) multiplicity
(c) x − c
4. Rational Zeros: If f(x) = an xn + an−1 xn−1 + ... + a1 x + a0 is a polynomial function s.t. ai ∈ Z, an ̸= 0, and f has
p
a zero that is a rational number, say
with p, q ∈ Z, then p must divide
a.
and q must divide
b.
.
q
(a) a0
1
(b) an
5. Standard form of a circle: The equation of a circle of radius r and center (h, k) is
(x − h)2 + (y − k)2 = r2
6. Explain why the vertical line test is a valid approach for testing whether a relation is a function of x.
The definition of a function requires that for each x in the domain of the function there is exactly one y in the range s.t.
f(x) = y. Thus a vertical line passing through x intersects the graph of the function exactly once at the point (x, y).
If there exists two distinct y’s in the range of f then f would not be a function by definition, and a vertical line
passing
through x would intersect the graph of f more than once. Hence the vertical line test is a suitable test to check
whether or not a relationship is a function of x.
7. Explain why the horizontal line test is a valid approach for testing whether a function is 1-1.
Using the same reasoning as problem 6 switch the roles of x and y and replace vertical line with horizontal line
and the test now follows.
2
Main Problems
Directions: Show all work and read the directions carefully.
8. Given the function y = f(x) defined by the graph on the left above, write the transformation necessary on the left
graph to produce the graph on the right in functional form.
Solution. y = f(−(x + 1)) + 1
9. Given the circle −7y + x2 + y2 =
51
4
(a) (6 pts) Write the equation for the circle in standard form.
Solution. Using the technique of completing the square we get the following equation:
51
4
51 49
=
+
4
4
51 49
+
=
4
4
( )2
10
=
2
−7y + x2 + y2 =
49
(x − 0)2 + y2 − 7y +
4
(
)2
7
(x − 0)2 + y −
2
(
)2
7
(x − 0)2 + y −
2
(b) What is the center and radius?
(
)
( )
7
10
Solution. ctr = 0,
and rad =
2
2
3
10. Use the following graph for parts a, b.
(a) Is y a function of x assuming a domain as shown in the graph above? If yes state a theorem to support your
answer. If not give a counterexample.
Solution. No, take as a counter example to the definition of a function the points (0, 0) and (0, 1)
(b) Is x a function of y assuming a domain as shown in the graph above? If yes state a theorem to support your
answer. If not give a counterexample.
Solution. Yes it passes the horizontal line test.
11. Define v(x) =
√
6 − 2x and h(x) = |2x + 3|
(a) (10 pts) Determine the domain of (vh)(x)
Solution. Recall the domain of the sum/difference/product/division of two functions is the intersection of
their respective domains (if division then we avoid dividing by zero of course). Thus since the domain of v is
the set of all x s.t. 6 − 2x ≥ 0 or x ≤ 3, and the domain of h is all real numbers we have the domain of (vh)(x)
as the intersection of (−∞, 3] and (−∞, ∞) or (−∞, 3]
(b) Determine the domain of (v ◦ h)(x)
Solution. Since the domain of h is all real numbers we need only find the set of all x s.t. h(x) = |2x + 3| is in
the domain of v or the set of all x s.t. |2x + 3| ≤ 3. We solve this inequality as follows
|2x + 3| ≤ 3
⇒
2x + 3 ≤ 3
∩
2x + 3 ≥ −3
⇒
∩
x ≥ −3
x≤0
⇒ [−3, 0]
4
12. Write a polynomial function that could represent the following graph given that the degree of the function is 4.
Solution. Using our knowledge of the properties of the multiplicities of zeros of a polynomial on its graph we know
that the multiplicity of −2 must be odd, 1 must be even, and 3 must be odd. Since the multiplicites must add up
to 4 we have that an odd + even + odd = 4 resulting in the only solution 1 + 2 + 1 = 4. Thus utilizing the factor
theorem we know that (x + 2), (x − 1)2 , and (x − 3) must be factors of the polynomial, p(x), in question. We need
only account for the constant out front by utilizing the y-int as follows:
p(x) = A(x + 2)(x − 1)2 (x − 3) and since p(0) = 1
⇒ 1 = A(0 + 2)(0 − 1)2 (0 − 3)
⇒ 1 = A(−6)
1
⇒A=−
6
1
Hence p(x) = − (x + 2)(x − 1)2 (x − 3)
6
13. Find the zeros(2pts), state there mulitiplicities(2pts), the x, y intercepts (4pts), holes(2pts), vertical asymptotes(2pts),
slant/horizontal asymptote(4pts), end behavior(2pts), domain(2pts), and sketch a graph(6pts)(be sure to label all
the properties x/y int, v-asyms, holes etc.) for the rational function defined by
f(x) =
2x2 − 3x + 1
x3 − 2x2 − 5x + 6
Solution. The numerator factors to 2x2 − 3x + 1 = (2x − 1)(x − 1) while the potential zeros of the denominator
1 −2 −5
6
are ±1, ±2, ±3 and ±6. Checking 1 we see that 1
1
3
2
1
−1
−1
−6
− 6 and thus we can factor the denominator
2
0
as follows x − 2x − 5x + 6 = (x − 1)(x − x − 6) = (x − 1)(x + 2)(x − 3). This leaves us with a factored form of
f(x) =
(2x − 1)(x − 1)
(x − 1)(x + 2)(x − 3)
This implies that the
5
1
with a multiplicity of one
2
(
)
1
x-int of f is
,0
2
(
)
1
y-int of f is 0,
6
(
)
1
hole of f is 1, −
6
vertical asymptotes of f are x = −2 and x = 3.
• zero of f is
•
•
•
•
• domain of f is R \ {1, −2, 3} or in interval notation (−∞, −2) ∪ (−2, 1) ∪ (1, 3) ∪ (3, ∞)
• horizontal asymptote of f is y = 0 since the degree of the denominator is larger than that of the numerator
(3 > 2), and
• the end behavior is thus as x → ∞ we have f(x) → 0 and as x → −∞ we also have f(x) → 0.
The graph of f (imagine the labels on the graph as you will be required to do so on your final) is then
14. Find all values of x s.t. e2x − 6ex − 16 = 0.
2
Solution. Observe that e2x − 6ex − 16 = (ex ) − 6 (ex ) − 16 and thus this is a quadratic in disguise as you can see
2
if you substitute say a = ex to get (ex ) − 6 (ex ) − 16 = a2 − 6a − 16, which factors to (a − 8)(a + 2). Setting to
zero and solving we get a = 8 and a = −2 re-substituting a = ex we solve for x by solving the equations ex = 8 and
ex = −2. The first gives us x = ln(8) while the second has no solution since -2 is not in the range of ex .
(
15. Show that
ln
)
√
)
(
√
c + c2 − x2
√
− 2 ln c + c2 − x2 + 2 ln(x) = 0
c − c2 − x2
6
Solution. One way to show this is as follows. Observe that
(
)
(
)
√
√
[(
(
)
)2 ]
√
√
2 − x2
c + c2 − x 2
c
+
c
√
√
ln
− 2 ln c + c2 − x2 + 2 ln(x) = ln
− ln c + c2 − x2
+ ln(x2 )
c − c2 − x 2
c − c2 − x 2


√
2
2
c+ c −x


2
= ln  (
)2 (
)  + ln(x )
√
√
c + c2 − x2
c − c2 − x2


(
)
√
2
2 − x2
x
c
+
c


= ln  (
)2 (
)
√
√
2
2
2
2
c+ c −x
c− c −x


2
x
)(
)
= ln  (
√
√
c + c2 − x2 c − c2 − x2
(
)
x2
= ln
(c2 − (c2 − x2 ))
( 2)
x
= ln
x2
= ln(1)
=0
as desired.
16. Let a, b ∈ R s.t. a, b ̸= 0. If (a + bi) is a zero of a quadratic function f (that has no vertical stretching), write out
the standard form of the quadratic function f (i.e. your answer should not be in factored form).
Solution. By the conjugate pairs theorem we know that (a − bi) must also be a zero. By the factor theorem we
know that f(x) = A(x − (a + bi))(x − (a − bi)), and since there is no stretching or reflections we know that A = 1
and thus
f(x) = (x − (a + bi))(x − (a − bi))
= x2 − x(a + bi) − x(a − bi) + (a + bi)(a − bi)
= x2 − 2ax + a2 + b2
17. Technetium-99, (T99 , is a radionuclide used widely in nuclear medicine. (T99 is combined with another substance
that is readily absorbed by a targeted body organ. Then, special cameras sensitive to the gamma rays emitted by
the technetium are used to record pictures of the organ. Suppose that a technician prepares a sample of (T99 to
image the heart of a patient suspected of having had a mild hear attack.
(a) At noon, the patient is given 10 mCi of (T99 . If the half-life of (T99 is 6hr, write a funciton of the form
Q(t) = Q0 ekt to model the radioactivity level Q(t) after t hours.
10
1
Solution. Utilizing the half life information we see that
= 10ek6 or that
= ek6 , which implies that
2
2
( −1 )
( )
ln 2
1
ln (2)
6k = ln
⇒k=
=−
and thus we have
2
6
6
Q(t) = 10e
−
ln (2)
t
6
(b) At what time will the level of radioactivity reach 3 mCi? Round to the nearest tenth of an hour.
7
Solution. Solving the equation Q(t) = 3 we get
ln (2)
t
6
3 = 10e
−
ln (2)
t
−
3
6
⇒
=e
( 10
)
3
ln (2)
⇒ ln
=−
t
10
6
( )
3
6 ln
10
⇒t=−
hrs
ln(2)
≈ 10.4hrs
8
in exact form, which is