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 + 1x 21). What can you say about the sign of this derivative? In view of (a)–(c) above, you should get conclusions that are in contradiction 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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 Fall '06
 McAdam
 Calculus, Fractions, Guoliang Wu

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