# HW6sol - M408C Homework 6 Solution Instructor Guoliang Wu 1...

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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?

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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...
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