# HW3 - momin(rrm497 Homework 03 cheng(58520 5 1 This...

This preview shows pages 1–4. Sign up to view the full content.

momin (rrm497) – Homework 03 – cheng – (58520) 1 This print-out should have 22 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A function h has graph -4 -3 -2 -1 0 1 2 3 2 - 2 2 - 2 on ( - 4 , 4). If f is defined on ( - 4 , 4) by f ( x ) = - 1 , - 4 < x < - 3 , integraldisplay x 3 h ( t ) dt, - 3 x < 4 , which of the following is the graph of f ? 1. -4 -3 -2 -1 0 1 2 3 2 - 2 2 - 2 - 4 correct 2. -4 -3 -2 -1 0 1 2 3 2 - 2 2 4 - 2 3. -4 -3 -2 -1 0 1 2 3 2 - 2 2 - 2 4. -4 -3 -2 -1 0 1 2 3 2 - 2 2 4 - 2 5. -4 -3 -2 -1 0 1 2 3 2 - 2 2 - 2 Explanation:

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

View Full Document
momin (rrm497) – Homework 03 – cheng – (58520) 2 Since f ( - 3) = integraldisplay 3 3 h ( t ) dt = 0 , two of the five graphs can be eliminated im- mediately. On the other hand, by the Fun- damental Theorem of Calculus, f ( x ) = h ( x ) on ( - 3 , 4); in particular, the critical points of f occur at the x -intercepts of the graph of h . As these x -intercepts occur at - 2 , 0 , 1, this eliminates a third graph. Thus the remain- ing two possible graphs for f both have the same critical points and to decide which one is the graph of f we can use the first derivative test because the graph of f will have a local maximum at an x -intercept of the graph of h where it changes from positive to negative values, and a local minimum at an x -intercept where h changes from negative to positive val- ues. Consequently, the graph of f must be -4 -3 -2 -1 0 1 2 3 2 - 2 2 - 2 - 4 keywords: 002 10.0 points If the function F is defined by F ( x ) = d dx parenleftBig integraldisplay x 4 1 8 t 6 dt parenrightBig , determine the value of F (1). Correct answer: 32. Explanation: By the Fundamental Theorem of Calculus, integraldisplay x 4 1 8 t 6 dt = bracketleftBig 8 7 t 7 bracketrightBig x 4 1 = 8 7 ( x 28 - 1) . Thus F ( x ) = d dx parenleftBig 8 7 ( x 28 - 1) parenrightBig = 32 x 27 Consequently, F (1) = 32. 003 10.0 points If f is a continuous function such that integraldisplay x 0 f ( t ) dt = 8 x 2 x 2 + 3 , find the value of f (1). 1. f (1) = 11 25 2. f (1) = 2 5 3. f (1) = 8 25 correct 4. f (1) = 12 25 5. f (1) = 9 25 Explanation: By the Fundamental Theorem of Calculus, d dx parenleftBig integraldisplay x 0 f ( t ) dt parenrightBig = f ( x ) . So by the Quotient Rule, f ( x ) = d dx parenleftBig 8 x 2 x 2 + 3 parenrightBig = 24 - 16 x 2 (2 x 2 + 3) 2 . In this case, f (1) = 8 25 .
momin (rrm497) – Homework 03 – cheng – (58520) 3 keywords: indefinite integral, Fundamental Theorem Calculus, FTC, function value, Quo- tient Rule, rational function, 004 10.0 points Determine F ( x ) when F ( x ) = integraldisplay x 4 4 sin t t dt . 1. F ( x ) = - 4 cos x x 2. F ( x ) = 2 cos( x ) x 3. F ( x ) = 2 sin( x ) x correct 4. F ( x ) = 4 sin x x 5. F ( x ) = - 2 sin( x ) x 6. F ( x ) = - 4 cos( x ) x 7. F ( x ) = 2 sin x x 8. F ( x ) = - 4 cos x x Explanation: By the Fundamental Theorem of Calculus and the Chain Rule, d dx parenleftBig integraldisplay g ( x ) a f ( t ) dt parenrightBig = f ( g ( x )) g ( x ) . When F ( x ) = integraldisplay x 4 4 sin t t dt , therefore, F ( x ) = 4 sin( x ) x parenleftBig d dx x parenrightBig .

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

View Full Document
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