AnswersPrecalculusApplying Vectors in the Plane

Finding Limits Answers

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1
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Consider the piecewise function.

A
Limit of f (x) = 2 as x approaches 1 minus. Limit of f (x) = 9 as x approaches 1 plus.
B
Limit of f (x) = 9 as x approaches 1 minus. Limit of f (x) = 2 as x approaches 1 plus.
C
Limit of f (x) = negative 4 as x approaches 1 minus. Limit of f (x) = negative 4 as x approaches 1 plus.
D
Limit of f (x) D N E as x approaches 1 minus. Limit of f (x) D N E as x approaches 1 plus.
2
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Review the graph of the function f(x).

Question illustration
A
The function is only left-continuous at x = 1.
B
The function is only right-continuous at x = 1.
C
The function is both left-continuous and right-continuous at x = 1.
D
The function is neither left-continuous nor right-continuous at x = 1.
3

Which function has a removable discontinuity at x = 7?

A
f (x) = StartFraction 3 Over x minus 7 EndFraction
Option A
B
f (x) = StartFraction x squared minus 10 x + 21 Over x minus 7 EndFraction
Option B
C
f (x) = StartLayout Enlarged left-brace first row negative x, x less-than 7 second row x + 1, x greater-than-or-equal-to 7 EndLayout
Option C
D
f (x) = StartLayout Enlarged left-brace first row negative StartFraction x Over x minus 7 EndFraction, x less-than 7 second row x + 1, x greater-than-or-equal-to 7 EndLayout
Option D
4

Based on the extreme value theorem, what is the maximum value of f(x) = –x2 + 6x over the interval [1, 4]?

A
3
B
5
C
8
D
9
5

Consider the function .Which statement explains the continuity of the function at x = –1?

Question illustration
A
Because f(x) = 4 and f(–1) = 4, it follows that f(x) = f(–1). Therefore, the function is continuous at x = –1.
B
Because f(x) = –4 and f(–1) = –4, it follows that f(x)= f(–1). Therefore, the function is continuous at x = –1.
C
Because f(x) = 4 and f(–1) = –4, it follows that f(x) ≠ f(–1). Therefore, the function is not continuous at x = –1.
D
Because f(x) = –4 and f(–1) = 4, it follows that f(x) ≠ f(–1). Therefore, the function is not continuous at x = –1.
6

Review the graph of a piecewise function.

Question illustration
A
–2
B
0
C
1
D
3
7

For which function and interval can the extreme value theorem be applied?

A
over the interval [3, 5]
Option A
B
over the interval (6, 8)
Option B
C
f(x) = 4 tan(πx) over the interval [0, 1]
D
f(x) = 3ex + 1 over the interval (1, 3)
8

Which function is continuous at x = –4?

A
g (x) = StartFraction x Over (x + 4) squared EndFraction
Option A
B
g (x) = StartFraction (x + 8) (x + 4) Over x + 4 EndFraction
Option B
C
g (x) = StartLayout enlarged left-brace first row x squared, x less-than negative 4 second row negative 4 x, x greater-than-or-equal-to negative 4 EndLayout
Option C
D
g (x) = StartLayout Enlarged left-brace first row x squared minus 4, x less-than negative 4 second row 5 x, x greater-than-or-equal-to negative 4 EndLayout
Option D
9

Review the graph of a piecewise function.

Question illustration
A
1
B
0 and 1
C
–2 and 1
D
–2, 0, and 1
10

Review the graph of function f(x).

Question illustration
A
The extreme value theorem applies, and f(x) has both a minimum and a maximum.
B
The extreme value theorem applies, and f(x) has a maximum but not a minimum.
C
The extreme value theorem does not apply, and f(x) does not have a minimum or a maximum.
D
The extreme value theorem does not apply, and f(x) has a maximum but not a minimum.
11

Review the graph of piecewise function g(x).

Question illustration
A
Option
Option A
B
Option
Option B
C
Limit of g (x) = 0 as x approaches negative 3 minus. Limit of g (x) = negative 2.5 as x approaches negative 3 plus.
Option C
D
Limit of g (x) = negative 2.5 as x approaches negative 3 minus. Limit of g (x) = 0 as x approaches negative 3 plus.
Option D
12

Using the extreme value theorem, which graph has a maximum of 5 and minimum of –3?

A
On a coordinate plane, a curve starts at closed circle (negative 1, negative 3) and curves up and approaches the y-axis in quadrant 2. A curve starts at closed circle (2, 5) and curves up and approaches x = 1 in quadrant 1.
B
On a coordinate plane, a curve starts at closed circle (negative 2, negative 3), curves up to (1, 0.5), curves down through (0, negative 1) to (0.5, negative 1.75), and curves up to closed circle (2, 5).
C
On a coordinate plane, a line starts at closed circle (negative 3, negative 3) and goes to closed circle (5, 1).
D
On a coordinate plane, a curve starts at closed circle (negative 4, negative 3), curves up to (1.5, 7.5) and curves down to closed circle (4, 5).
13

Function f(x) has a jump discontinuity at x = 8. Which statement identifies possible one-sided limits at x = 8?

A
DNE and
Option A
B
DNE and
Option B
C
= 85 and
Option C
D
= 85 and
Option D
14

Function has a discontinuity at x = –3. What are and ?

Question illustration
A
Limit of f (x) = infinity as x approaches negative 3 minus. Limit of f (x) = infinity as x approaches negative 3 plus.
B
Limit of f (x) = negative infinity as x approaches negative 3 minus. Limit of f (x) = infinity as x approaches negative 3 plus.
C
Limit of f (x) = infinity as x approaches negative 3 minus. Limit of f (x) = negative infinity as x approaches negative 3 plus.
D
Limit of f (x) = negative infinity as x approaches negative 3 minus. Limit of f (x) = negative infinity as x approaches negative 3 plus.

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