Unit Test — Unit test Answers

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Review the graph of a piecewise function.

Question illustration
A
jump discontinuity at x = –2; mixed discontinuity at x = 0
B
jump discontinuity at x = –2; infinite discontinuity at x = 0
C
endpoint discontinuity at x = –2; mixed discontinuity at x = 0
D
endpoint discontinuity at x = –2; infinite discontinuity at x = 0
3

Which function is continuous at x = 18?

A
f (x) = StartFraction (x minus 18) squared Over x EndFraction
Option A
B
f (x) = StartFraction x squared minus 17 x minus 18 Over x minus 18 EndFraction
Option B
C
f (x) = tangent (StartFraction pi Over 36 EndFraction x)
Option C
D
f (x) = StartLayout Enlarged left-brace first row x squared, x not-equals 18 second row 36, x = 18 EndLayout
Option D
4

Review the graph of f(x).

Question illustration
A
Around . Around x = 3, the function behaves in the same manner as around x = –3.
Option A
B
Around . Around x = 3, the function behaves in the same manner as around x = –3.
Option B
C
Around . Around x = 3, the function behaves in the opposite manner as around x = –3.
Option C
D
Around . Around x = 3, the function behaves in the opposite manner as around x = –3.
Option D
5

Which statement describes the sequence defined by ?

Question illustration
A
The sequence diverges.
B
The sequence converges to a limit of 0.
C
The sequence converges to a limit of ∞.
D
The sequence converges to a limit of –∞.
6

What is ?

Question illustration
A
Negative one-sixth
Option A
B
0
C
One-sixth
Option C
D
DNE
7

Review the table of values for function g(x).xg(x)2.9–1.552.99–1.582.999–1.593.0011.593.011.583.11.55

A
The limits = –1.6 and = 1.6. Both and exist, so exists.
Option A
B
The limits = 1.6 and = –1.6. Both and exist, so exists.
Option B
C
The limits = –1.6 and = 1.6. Because ≠ , does not exist.
Option C
D
The limits = 1.6 and = –1.6. Because ≠ , does not exist.
Option D
9

What is the horizontal asymptote of

Question illustration
A
y = 1
B
y = 5
C
y = 7
D
y = 12
11

Which graph has the same end behavior as

Question illustration
A
On a coordinate plane, a graph approaches the x-axis in quadrant 2 and then increases up to (negative 0.5, 7.5), down to (0, 6), up to (0.5, 6.3) and then curves down and approaches the x-axis in quadrant 1.
Option A
B
On a coordinate plane, a graph curves down through (negative 1, 6) to (0.2, 1.8), has vertex (1, 0), and then curves up through (3, 8).
Option B
C
On a coordinate plane, a graph curves up and approaches x = negative 2, curves down to (negative 1, 3), curves up to (0, 5), and then curves down and approaches x = 2 in quadrant 4.
Option C
D
On a coordinate plane, a curve approaches x = 1 in quadrant 1 and curves down through (0, 0) and (negative 3, negative 2). Another curves approaches x = 1 in quadrant 4 and curves up through (2, negative 2.2) and (4, negative 2.4).
Option D
12

Consider the series Which expression defines Sn?

Question illustration
A
Limit of (one-fourth) Superscript n Baseline as n approaches infinity
Option A
B
Limit of (1 minus (one-fourth) Superscript n Baseline) as n approaches infinity
Option B
C
Limit of one-third (1 minus (one-fourth) Superscript n Baseline) as n approaches infinity
Option C
D
Limit of one-third (one-fourth) Superscript n Baseline as n approaches infinity
Option D
13

Over the closed interval [3, 8], for which function can the extreme value theorem be applied?

A
h (x) = StartFraction negative 2 Over 5 (x minus 4) squared EndFraction
Option A
B
h (x) = StartFraction (x minus 5) (x minus 1) Over x squared minus 25 EndFraction
Option B
C
h (x) = StartLayout Enlarged left-brace first row StartFraction 9 x Over 10 minus x EndFraction, x less-than 4 second row x + 2, x greater-than-or-equal-to 4 EndLayout
Option C
D
h (x) = StartLayout Enlarged left-brace first row negative x, x less-than 5 second row x squared minus 20, x greater-than-or-equal-to 5 EndLayout
Option D
15

Which statement correctly uses limits to determine a vertical asymptote of

Question illustration
A
There is a vertical asymptote at x = 5, because
Option A
B
There is a vertical asymptote at x = 5, because
Option B
C
There is a vertical asymptote at x = –5, because
Option C
D
There is a vertical asymptote at x = –5, because
Option D
16

Consider the sequence Which statement describes the sequence?

Question illustration
A
The sequence diverges.
B
The sequence converges to 0.
C
The sequence converges to 1.
D
The sequence converges to ∞.
18

Given . What is ?

Question illustration
A
–2
B
–1
C
1
D
2
19

Review the graph of the function f(x).

Question illustration
A
Limit of f (x) = 2 as x approaches 1 minus. Limit of f (x) D N E as x approaches 1 plus.
Option A
B
Limit of f (x) = negative 2 as x approaches 1 minus. Limit of f (x) D N E as x approaches 1 plus.
Option B
C
Limit of f (x) D N E as x approaches 1 minus. Limit of f (x) D N E as x approaches 1 plus.
Option C
D
Limit of f (x) D N E as x approaches 1 minus. Limit of f (x) = negative 2 as x approaches 1 plus.
Option D
20

What is ?

Question illustration
A
0
B
1
C
3
D
DNE

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