MATH 1450 Spring 2010 Quiz 5 Solutions

MATH 1450 Spring 2010 Quiz 5 Solutions - bottomed out and...

Info icon This preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
MATH 1450, QUIZ 5 SOLUTIONS, 18 FEB., 2010 (1) Fill in the blanks: The expression lim h 0 1 / (1 + h ) 2 - 1 h represents the of the function f ( x ) = , at the point x = . The expression is the derivative of the function f ( x ) = 1 x 2 , at the point x = 1. ( 1) Fill in the blanks: The expression lim h 0 1 / (2 + h ) 2 - 1 / 2 h represents the of the function f ( x ) = , at the point x = . The expression is the derivative of the function f ( x ) = 1 x 2 , at the point x = 2. (2) Sketch the graph of a function y = f ( x ), continuous on the interval [0 , 4] and satisfying the conditions: f ( x ) < 0 for 0 < x < 1 f ( x ) > 0 for 1 < x < 2 f ( x ) = 0 for 2 < x < 4 As you draw the graph west-to-east, starting at x = 2, the graph slopes downhill until you reach x = 1. At this point the graph has
Image of page 1

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

View Full Document Right Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: bottomed out and starts going uphill. When you reach x = 2, level o± and proceed due east until x = 4. ( 2) Sketch the graph of a function y = f ( x ), continuous on the interval [0 , 4] and satisfying the conditions: • f ′ ( x ) > 0 for 0 < x < 1 • f ′ ( x ) < 0 for 1 < x < 2 • f ′ ( x ) = 0 for 2 < x < 4 As you draw the graph west-to-east, starting at x = 2, the graph slopes uphill until you reach x = 1. At this point the graph has 1 topped out and starts going downhill. When you reach x = 2, level of and proceed due east until x = 4....
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