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

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

View Full Document Right Arrow Icon
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:
Image of page 1

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

View Full Document Right Arrow Icon
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 .
Image of page 2
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 .
Image of page 3

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

View Full Document Right Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    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.

    Student Picture

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

  • Left Quote Icon

    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.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    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.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern