Calculus

# Calculus - Create assignment 57950 Final 1 Dec 03 at 4:29...

This preview shows pages 1–5. Sign up to view the full content.

Create assignment, 57950, Final 1, Dec 03 at 4:29 pm 1 This print-out should have 109 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. CalC1b05 47:02, calculus3, multiple choice, > 1 min, wording-variable. 001 Sketch the graph of the function f ( x ) = m ( x - 3) + 4 for m = - 1 , 0 , 1. 1. 4 8 - 4 4 8 - 4 x y cor- rect 2. 4 8 - 4 4 8 - 4 x y 3. 4 8 - 4 4 8 - 4 x y 4. 4 8 - 4 4 8 - 4 x y 5. 4 8 - 4 4 8 - 4 x y Explanation: Since each function f is linear, its graph will be a straight line. Furthermore, these lines will pass through (3,4) and have respective slope m = - 1 , 0 , 1. Consequently, the three graphs are given by

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Create assignment, 57950, Final 1, Dec 03 at 4:29 pm 2 4 8 - 4 4 8 - 4 x y CalC1c27 47:02, calculus3, multiple choice, > 1 min, wording-variable. 002 Which one of the following is the graph of the function f ( x ) = sin | 3 x | ? 1. 1 - 1 1 2 3 - 1 - 2 - 3 correct 2. 1 - 1 1 2 3 - 1 - 2 - 3 3. 1 - 1 1 2 3 - 1 - 2 - 3 4. 1 - 1 1 2 3 - 1 - 2 - 3 5. 1 - 1 1 2 3 - 1 - 2 - 3 Explanation: Note: The first full cycle of the sin(3 x ) function occurs when 3 x = 2 π ; i.e. , when x = 2 π 3 = 2 . 0944. Now the graph of the function f ( x ) = sin 3 x , without the absolute value, is
Create assignment, 57950, Final 1, Dec 03 at 4:29 pm 3 1 - 1 1 2 3 - 1 - 2 - 3 However, because | - x | = | x | , | x | = x ( x 0) , the function f ( x ) = sin | 3 x | is an even function ( f ( - x ) = f ( x )), so the graph to the left of the y -axis is the mirror image of the graph of sin 3 x to the right of the y -axis. Consequently, the graph of f is: 1 - 1 1 2 3 - 1 - 2 - 3 CalC1a26s 47:02, calculus3, multiple choice, > 1 min, normal. 003 Find the natural domain of the function g ( u ) = u - 7 - u . 1. nat. domain = [0 , 7] correct 2. nat. domain = (0 , 7) 3. nat. domain = [0 , ) 4. nat. domain = (0 , 7] 5. nat. domain = [0 , 7) Explanation: The natural domain of g consists of all val- ues of u for which both of the functions g 1 ( u ) = u, g 2 ( u ) = 7 - u are defined. But g 1 is defined for all u 0, while g 2 is defined for all u 7. Consequently, natural domain = [0 , 7] . CalC1a35a 47:02, calculus3, multiple choice, > 1 min, wording-variable. 004 Which function has 1 2 3 4 5 6 1 2 3 4 5 as its graph on [ - 1 , 7]? 1. f ( x ) = 3 2 + 1 x - 2 + 1 | x - 2 | correct 2. f ( x ) = 3 2 - 1 x - 2 - 1 | x - 2 | 3. f ( x ) = 1 x - 2 + 1 | x - 2 | - 3 2 4. f ( x ) = 3 2 + 1 x - 2 - 1 | x - 2 |

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Create assignment, 57950, Final 1, Dec 03 at 4:29 pm 4 5. f ( x ) = 3 2 - 1 x - 2 + 1 | x - 2 | Explanation: Since | x - 2 | = x - 2 , x 2, 2 - x, x < 2, we see that 1 x - 2 + 1 | x - 2 | = 2 x - 2 , x > 2, 0 , x < 2 . Notice that the value x = 2 has to ex- cluded because it does not belong to the nat- ural domain of 1 / ( x - 2) or of 1 / | x - 2 | . In particular, the positive x -axis is a horizontal asymptote, while the line x = 2 for y > 0 is a vertical asymptote of the graph 1 2 3 4 5 6 1 2 3 4 5 of the function defined by g ( x ) = 1 x - 2 + 1 | x - 2 | . The circle indicates that g ( x ) is not defined at x = 2. Since the given graph is just the graph of g shifted vertically upwards by 3 2 units, the given graph is that of the function f ( x ) = 3 2 + 1 x - 2 + 1 | x - 2 | .
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern