# HW6 - x 1< x 2 1 for all x> 0(a Assume(for the sake of...

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M408C, Homework 6 Instructor: Guoliang Wu 1 Quest – Deadline: Wed. 10 / 6 4:00 PM Please log onto Quest https://quest.cns.utexas.edu , and answer the ques- tions online. I f your response is a fraction , convert it into a decimal . Q uest does not recognize fractions ! If you lose points because of this before, and you really want to get the points back, you can drop by during any of my o ce hours and we can ﬁx that (I just learned how to do that). 2 Problems from textbook – Deadline: Thur. 10 / 7, Discussion Section Please explain your work! Correct answers alone are not enough. 1. Find the absolute maximum and absolute minimum of the function f ( x ) = x 4 - x 2 , on [ - 1 , 2] . 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,
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Unformatted text preview: 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-? (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 ( x ) = d dx ( √ x + 1-x 2-1). What can you say about the sign of this derivative? In view of (a)–(c) above, you should get conclusions that are in contra-diction 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”.) 1...
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