HW6sol - tion to each other The only place that can go...

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

View Full Document Right Arrow Icon
M408C, Homework 6 — Solution Instructor: Guoliang Wu 1. Find the absolute maximum and absolute minimum of the function f ( x ) = x p 4 - x 2 , on [ - 1 , 2] . Solution. f 0 ( x ) = p 4 - x 2 + x - 2 x 2 4 - x 2 = 4 - 2 x 2 4 - x 2 So f 0 ( x ) = 0 4 - 2 x 2 = 0 x 2 = 2 x = ± 2 . Note that - 2 - 1 . 414 is not in [ - 1 , 2]. So 2 is the only critical point. We evaluate f at 2 , - 1 and 2, f ( 2) = 2 p (4 - 2) = 2 , f ( - 1) = - 3 , f (2) = 0 . So the absolute maximum is f ( 2) = 2 and the absolute minimum is f ( - 1) = - 3. 2. Let f ( x ) = x + 1 - x 2 - 1 for x 0. The following sequence of questions will help you to understand why f ( x ) < 0 (equivalently, x + 1 < x 2 + 1) for all x > 0. (a) Assume (for the sake of contradiction) that f ( b ) 0 at some point b > 0. What can you say about the sign of the value f ( b ) - f (0) b - 0 ? (b) Can you apply the Mean Value Theorem to f ( x ) on [0 , b ]? Why? And what can you conclude from it, assuming f ( b ) 0? (c) Now compute the derivative f 0 ( x ) = d dx ( x + 1 - x 2 - 1). What can you say about the sign of this derivative?
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: tion to each other. The only place that can go wrong the assumption that f ( b ) ≥ 0 at some point b > 0. Therefore, f ( x ) < 0 for all x > 0. (This argument is called “proof by contradiction”.) Solution. (a) Since we assume f ( b ) ≥ , f ( b )-f (0) b-= f ( b ) b ≥ . 1 (b) Since the function f ( x ) is continuous on [0 ,b ] and differentiable on (0 ,b ), we can apply the MVT. From MVT, we conclude that there exists c in (0 ,b ) such that f ( c ) = f ( b )-f (0) b-= f ( b ) b ≥ (c) f ( x ) = 1 2 √ x +1-1 2 = 2(1-√ x +1) 2 √ x +1 < , for x > 0. But f ( c ) ≥ 0 from (b), where c in (0 ,b ), so we have a contradiction. 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