review 1 ans - Lescure Etienne Review 1 Due 10:00 pm Inst...

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

View Full Document Right Arrow Icon
Lescure, Etienne – Review 1 – Due: Dec 10 2007, 10:00 pm – Inst: Diane Radin 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points Find the derivative of g when g ( x ) = x 4 cos x . 1. g 0 ( x ) = x 4 (3 cos x - sin x ) 2. g 0 ( x ) = x 3 (4 cos x - x sin x ) correct 3. g 0 ( x ) = x 3 (4 sin x - x cos x ) 4. g 0 ( x ) = x 3 (4 cos x + x sin x ) 5. g 0 ( x ) = x 4 (3 sin x - cos x ) 6. g 0 ( x ) = x 3 (4 sin x + x cos x ) Explanation: By the Product rule, g 0 ( x ) = x 4 ( - sin x ) + (cos x ) · 4 x 3 . Consequently, g 0 ( x ) = x 3 (4 cos x - x sin x ) . keywords: derivative, product rule, trigono- metric function 002 (part 1 of 1) 10 points Find the derivative of the function f ( x ) = 3 - 2 x x . 1. f 0 ( x ) = 6 - 2 x x 2 2. f 0 ( x ) = 6 - 2 x x 3 3. f 0 ( x ) = 4 - 3 x x 3 correct 4. f 0 ( x ) = 4 - 3 x x 2 5. f 0 ( x ) = 4 + 3 x x 3 Explanation: After simplification, 3 - 2 x x = 3 x - 2 x 2 . Thus by the Quotient Rule, f 0 ( x ) = 3 x 2 - 2 x (3 x - 2) x 4 . Consequently, f 0 ( x ) = 4 - 3 x x 3 . keywords: derivatives, quotient rule 003 (part 1 of 1) 10 points Find the derivative of f when f ( x ) = sin x 1 - 4 cos x . 1. f 0 ( x ) = cos x - 4 (1 - 4 cos x ) 2 correct 2. f 0 ( x ) = sin x - 4 1 - 4 cos x 3. f 0 ( x ) = cos x + 4 1 - 4 cos x 4. f 0 ( x ) = cos x + 4 (1 - 4 cos x ) 2 5. f 0 ( x ) = sin x - 4 cos 2 x (1 - 4 cos x ) 2
Image of page 1

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

View Full Document Right Arrow Icon
Lescure, Etienne – Review 1 – Due: Dec 10 2007, 10:00 pm – Inst: Diane Radin 2 6. f 0 ( x ) = cos x - 4 sin 2 x (1 - 4 cos x ) 2 Explanation: By the Quotient Rule, f 0 ( x ) = cos x (1 - 4 cos x ) - 4 sin x sin x (1 - 4 cos x ) 2 = cos x - 4 (cos 2 x + sin 2 x ) (1 - 4 cos x ) 2 . But cos 2 x + sin 2 x = 1, so f 0 ( x ) = cos x - 4 (1 - 4 cos x ) 2 . keywords: derivative, trigonometric function, quotient rule 004 (part 1 of 1) 10 points There is one point in the first quadrant at which the tangent line to the graph of y = 5 + x + x 2 - x 3 is horizontal. Find the y -coordinate of this point. 1. y = 6 correct 2. y = 8 3. y = 7 4. y = 9 5. y = 5 Explanation: The tangent line to the graph will be hori- zontal when dy dx = 1 + 2 x - 3 x 2 = (3 x + 1)(1 - x ) = 0 . The only solution of this for which x > 0 oc- curs at x = 1. But at x = 1 the corresponding value of y is y = 6. Since this value of y is positive, the only point in the first quadrant at which the tangent line is horizontal is the point P = (1 , 6) . keywords: horizontal tangent line, derivative, extrema, polynomial 005 (part 1 of 1) 10 points If f is a function defined on ( - 2 , 2) whose graph is 1 2 - 1 - 2 1 2 - 1 - 2 which of the following is the graph of its derivative f 0 ? 1. 1 2 - 1 - 2 1 2 - 1 - 2
Image of page 2
Lescure, Etienne – Review 1 – Due: Dec 10 2007, 10:00 pm – Inst: Diane Radin 3 2. 1 2 - 1 - 2 1 2 - 1 - 2 correct 3. 1 2 - 1 - 2 1 2 - 1 - 2 4. 1 2 - 1 - 2 1 2 - 1 - 2 5. 1 2 - 1 - 2 1 2 - 1 - 2 6. 1 2 - 1 - 2 1 2 - 1 - 2 Explanation: Since the graph on ( - 2 , 2) consists of straight lines joined at x = - 1 and x = 1, the derivative of f will exist at all points in ( - 2 , 2) except x = - 1 and x = 1, eliminating the answer whose graph contain filled dots at x = - 1 and x = 1. On the other hand, the graph of f (i) has slope - 1 on ( - 2 , - 1), (ii) has slope 2 on ( - 1 , 1), and (iii) has slope - 1 on (1 , 2).
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