QUIZ 1B - SOLUTIONS
(CHAPTER 1: FUNCTIONS)
MATH 141 – FALL 2016 – KUNIYUKI (60 POINTS TOTAL)
()
1) (5 points). Write the domain of f , where f x =
the form using parentheses and/or brackets.
x+2
, using interval form,
x−3
x + 2 is real ⇔ x + 2 ≥ 0 ⇔ x ≥ − 2 .
•
• Now, x − 3 ≠ 0 ⇔ x ≠ 3 , so 3 is the only exclusion from the domain based on the
denominator, which is not allowed to be 0.
( )
Here is a graph of the domain, Dom f :
( )
) (
)
In interval form, Dom f = ⎡⎣ − 2, 3 ∪ 3, ∞ .
2) (3 points). Find and box in the x-intercept(s) (if any) of the graph of
x 2 − 25
in the usual xy-plane. If there are none, write “NONE.”
x−5
y=
To find x-intercepts, we substitute (“plug in”) y = 0 . Find the real solutions of:
0=
x 2 − 25
.
x−5
x 2 − 25
. That is, we want the real zeros of f , where f x =
x−5
(
()
f x =0 ⇔
(x
⇔
()
)
( x − 5 ≠ 0)
( x ≠ 5)
x 2 − 25 = 0 and
2
)
− 25 = 0 and
The real zeros of x 2 − 25 are 5 and −5 . We can use the Quadratic Formula (QF) or:
Square root method
Factoring method
x 2 − 25 = 0
x 2 − 25 = 0
( x + 5)( x − 5) = 0
(
)
(
)
⎡ x + 5 = 0 or x − 5 = 0 ⎤
⎣
⎦
x = ±5 ( x = 5 or x = −5)
x = ±5 ⎡⎣ x = −5 or x = 5⎤⎦
However, “5” must be rejected, because it is not in the domain of f . We only take −5 .
x 2 = 25
(
)
The only x-intercept is at −5, 0 .
()
Here is the graph of y = f x .
3) (4 points total). Box in the appropriate answers.
a)
a) The graph of y = x 5 − 2x in the usual xy-plane is symmetric about …
b)
the x-axis
()
f (− x) = (− x)
the y-axis
the origin
(none of these)
( )
− 2 ( − x ) = − x + 2x = − ( x − 2x ) = − f ( x )
Therefore, the graph of y = f ( x ) is symmetric about the origin.
Let f x = x 5 − 2x . Dom f =  . The function f is odd because ∀x ∈ ,
5
5
5
()
b) If g t = t 2/3 + 5 , then g is …
even
odd
neither
Dom ( g ) =  . The function g is even because ∀t ∈ ,
g ( t ) = t 2/3 + 5 =
g (− t ) =
3
(− t )
(
2
3
t2 + 5 ⇒
+5 =
3
t 2 + 5 = t 2/3 + 5 = g ( t )
)
()
4) (6 points total). If the point 1, −5 lies on the graph of y = f x , where f
is a one-to-one function, what point must then lie on the graph of …
(
)
a) … y = f x − 1 + 4 ?
( 2, −1) , after the point (1, −5) is shifted 1 unit right and 4 units up.
()
b) … y = − f x − 2 ?
(1, 3) , because the point (1, −5) is reflected about the x-axis, thus
yielding the point (1, 5) ; this point is then shifted two units down, yielding
the point (1, 3) .
c) … y = f
−1
( x) ?
( −5, 1) , the reflection of the point (1, −5) about the line
We switch the x- and y-coordinates.
5) (2 points). Evaluate  − 7.1 . (This is the same as ⎢⎣ − 7.1⎥⎦ .)
− 8 ; the greatest integer (or floor) function rounds down.
− 8 is the greatest integer that does not exceed − 7.1 .
y = x.
6) Match the equations with their corresponding graphs by writing the
appropriate letters in the blanks. The x- and y-axes are not necessarily scaled the
same way. (6 points total)
1
The graph of y = is Graph __A__. (Hyperbola)
x
1
The graph of y = 2 is Graph __F__. (Volcano)
x
3
The graph of y = x is Graph __B__. (Lazy snake)
The graph of y = x 2/3 is Graph __D__. (Bird beak / Funnel / Tornado)
The graph of y = 16 − x 2 is Graph __E__. (Upper semicircle)
The graph of y = x is Graph __C__. (V graph)
Graph A
Graph B
Graph C
Graph D
Graph E
Graph F
Observe that A and B are graphs of odd functions, while C, D, E, and F are graphs of
even functions.
The domains of the functions corresponding to A, B, C, D, E, and F are, respectively,
− ∞, 0 ∪ 0, ∞ ,  ,  ,  , ⎡⎣ − 4, 4 ⎤⎦ , and − ∞, 0 ∪ 0, ∞ .
(
) (
)
(
) (
(
)
)( )
7) (2 points). Find functions g and f such that f  g x = 3 x 2 + 1 .
You may not use the identity function. Fill in the blanks:
g ( x ) = x 2 + 1, f ( u ) = 3 u . There are many other possibilities.
8) (2 points). Sketch the graph of x = − y 2 in the usual xy-plane.
9) Let s ( t ) = t 3 − 2t . Find the average rate of change of s from t = 1 to t = 4 .
Assume that t is time measured in seconds and s ( t ) is the position of a particle
measured in meters. Write the appropriate unit in your final answer.
Note: You are finding the average velocity of the particle between t = 1 second
and t = 4 seconds. (7 points)
s is a polynomial function, so the usual formula applies.
s ( 4 ) − s (1)
4 −1
⎡( 4 )3 − 2 ( 4 ) ⎤ − ⎡(1)3 − 2 (1) ⎤
56 − ⎡⎣ −1⎤⎦
57
meters
⎢
⎥⎦ ⎢⎣
⎥⎦
= ⎣
=
=
= 19
3
3
3
second
⎧⎪ x 2 + 1, − 2 ≤ x ≤ 1
10) f is the function defined piecewise by: f ( x ) = ⎨
. (13 pts.)
⎪⎩ x − 1, 1 < x < 5
a) Evaluate f ( − 2 ) . (1 point)
− 2 ≤ − 2 ≤ 1 , so use the top rule:
f (− 2) = (− 2) + 1 = 4 + 1 = 5 .
f ( x) = x2 + 1 ⇒
2
b) Evaluate f (1) . (1 point)
− 2 ≤ 1 ≤ 1 , so use the top rule:
f ( x) = x2 + 1 ⇒
f (1) = (1) + 1 = 1+ 1 = 2 .
2
c) Evaluate f ( 2 ) . (1 point)
1 < 2 < 5 , so use the bottom rule:
f ( x) = x − 1 ⇒
f (2) =
(2) − 1
=
1 = 1.
d) Graph y = f ( x ) on the grid below. Be as accurate as possible.
Clearly indicate whether endpoints are included or excluded. (6 points)
Locate endpoints on the graph (corresponding to the endpoints of the
subdomains) and determine whether they are included in (or excluded from) the
graph.
Rule
Left endpoints
Right endpoints
Shapes
x = −2
x =1
The parabola with
y = (− 2) + 1 = 5
(see Part a).)
Left endpoint: ( − 2, 5)
Included because of “ ≤ ”
(weak inequality).
x =1
2
x2 + 1
y=
x −1
(1) − 1 =
0 =0
( )
Left endpoint: 1, 0
Excluded because of “<”
(strict inequality).
y = (1) + 1 = 2
(see Part b).)
Right endpoint: (1, 2 )
Included because of “ ≤ ”
(weak inequality).
x=5
2
y=
(5) − 1 =
4 =2
( )
Right endpoint: 5, 2
Excluded because of “<”
(strict inequality).
equation y = x 2 + 1 is
the graph of y = x 2
shifted one unit up.
We take a piece of
this graph.
The graph of
y = x − 1 is the
“hook” graph of
y = x shifted one
unit to the right.
We take a piece of
this graph.
()
( )
From Part c), we see that f 2 = 1 , so the point 2, 1 is also on the graph.
e) Give the domain of the function f using interval form (the form with
parentheses and/or brackets). (2 points)
Dom ( f ) = ⎡⎣ − 2, 5) , the union of the given subdomains: ⎡⎣ − 2, 1⎤⎦ ∪ (1, 5) . The
()
domain corresponds to the x-coordinates “picked up” by the graph of y = f x .
f) Give the range of the function f using interval form (the form with
parentheses and/or brackets). (2 points)
Range ( f ) = ( 0, 5⎤⎦ . The range corresponds to the y-coordinates
()
“picked up” by the graph of y = f x .
()
11) Let f x = x . Simplify the difference quotient completely:
f ( x + h) − f ( x )
h
Hint: You will need to rationalize a numerator. (10 points)
∀h ≠ 0,
f ( x + h) − f ( x )
h
(
=
=
h
x+h − x
h
(
)⋅(
(
x +h −x
x+h + x
)
x+h − x
h
=
( Rationalize the numerator.)
) = ( x + h) − ( x )
x)
h( x + h + x )
2
x+h + x
x+h +
2
=
(1)
=
(
()
h
1
h
x+h + x
)
1
=
As h approaches 0, this approaches
x+h + x
1
2 x
( x + h) − ( x )
h
(
x+h + x
)
( h ≠ 0)
()
()
, which is f ′ x , the derivative of f x .