{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Quiz2s - (b(5 points Let f be the function f x = x 2-1 |...

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
MATH 1501 Quiz 2 August 31, 2009 No books or notes allowed. No laptop, graphic calculator or wireless devices allowed. Write clearly. Name:
Image of page 1

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

View Full Document Right Arrow Icon
MATH 1501 Quiz 2 August 31, 2009 1. (a) (5 points) Let f be the function f ( x ) = ( x - 1) 2 | x - 1 | . Can you define f for x = 1 so that f is continuous at 1. ( Hint: check the right and left limit separatly.) Solution: For x > 1 we have f ( x ) = ( x - 1) x - 1 | x - 1 | = x - 1 while for x < 1 f ( x ) = ( x - 1) x - 1 | x - 1 | = - ( x - 1) so that lim x 1 + f ( x ) = lim x 1 - f ( x ) = 0 Thus setting f (1) = 0 we have that f ( x ) is continuous at 1. Page 1 of 2
Image of page 2
MATH 1501 Quiz 2 August 31, 2009
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: (b) (5 points) Let f be the function f ( x ) = x 2-1 | x-1 | . Can you define f for x = 1 so that f is continuous at 1. Solution: For x > 1 we have f ( x ) = ( x + 1) x-1 | x-1 | = x + 1 while for x < 1 f ( x ) = ( x + 1) x-1 | x-1 | =-( x + 1) so that lim x → 1 + f ( x ) = 2 lim x → 1-f ( x ) =-2 Thus f ( x ) has a jump discontinuity at 1. There is no way to define f for x = 1 so that f is continuous at 1. Page 2 of 2...
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