{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Q4Solutions - Now since f 00< 0 for all x 6 = 0 x = 0 is...

Info icon This preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
MAC 2233 Quiz 4 Solutions July 1, 2009 You must show all work to receive credit 1. Let f ( x ) = 3 x 4 - 6 x 3 + x - 8. Find the open intervals where f is concave down. Solution It suffices to find all open intervals where f 00 ( x ) < 0. Begin by computing f 0 and f 00 . We have f 0 ( x ) = 12 x 3 - 18 x 2 + 1 f 00 ( x ) = 36 x 2 - x = 36 x ( x - 1) . The zeros of f 00 are x = 0 and x = 1. By computing f 00 ( - 1) , f 00 ( 1 2 ) and f 00 (2), we see that f 00 ( x ) < 0 on (0 , 1) and f 00 ( x ) > 0 on ( -∞ , 0) (1 , ). It follows that f is concave down on (0 , 1). 2. Find all inflection points of the function f ( x ) = x 4 7 . If there are none, explain why . Solution A routine computation shows that f 00 ( x ) = - 12 49 x 10 / 7 . By inspecting f 00 , we see that f 00 6 = 0 for any x in the domain of f . Also, we see that f 00 does not exist at x = 0, so x = 0 is the only candidate to be an inflection point.
Image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Now, since f 00 < 0 for all x 6 = 0, x = 0 is not an inflection point of f . 3. Consider the function f ( x ) = 2 x 3 + 3 x 2-12 x-4. The critical points of f are x 1 = 1 and x 2 =-2. Moreover, f (1) = 0 and f (-2) = 0. If f 00 ( x ) = 6(2 x + 1), classify each of x 1 and x 2 as local maximizers or local minimizers. Solution Since each of x 1 = 1 and x 2 =-2 are zeros of f , we need only com-pute the signs of f 00 ( x 1 ) and f 00 ( x 2 ). A routine calculation shows that f 00 (1) > 0 and f 00 (-2) < 0. Therefore, x 1 = 1 is a local minimizer of f and x 2 =-2 is a local maximizer of f ....
View Full 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