275_exam2_solution

# 2 5 suppose that the function f x is continuous on

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5. Suppose that the function f ( x ) is continuous on the open interval (0 , 1), and we also have lim x 0 + f ( x ) = 0, lim x 1 - f ( x ) = 1. (a) (Show that there is a continuous function g ( x ) on [0 , 1] for which g ( x ) = f ( x ) for every x (0 , 1). Deﬁne the function g ( x ) on [0 , 1] as follows: g ( x ) = 0 if x = 0 f ( x ) if 0 < x < 1 1 if x = 1 This is continuous for every x (0 , 1) (since it’s equal to f which is continuous by assumption). It is also right-continuous at x = 0 (since lim x 0 + f ( x ) = 0 = g (0)) and the same is true for x = 1. But this means that g is continuous on [0 , 1]. (b) Show that f is bounded in (0 , 1). The function g is continuous on [0 , 1] so by the Extreme Value Theorem it is bounded there: | g ( x ) | < M for some M for x [0 , 1]. But then | f ( x ) | < M is true for all x (0 , 1) since g ( x ) = f ( x ) there. (c) Is it always true that f has an absolute maximum in (0 , 1)? If yes, prove it, if not then give a counterexample. No. E.g. f ( x ) = x is a function like this and it does not have an absolute maximum in (0 , 1). (Since it’s strictly less than 1 and it can be bigger than 1 - ε for any ε > 0.) 6. Show that the equation x = cos( x 3 ) has at least one real solution. The function f ( x ) = x - cos( x 3 ) is continuous for every x 0 (right-continuous at x = 0). We have f (0) = - 1 and f (4) = 4 - cos(64) 2 - 1 = 1. Thus f (0) < 0 ,f (4) > 1 and by Bolzano’s theorem there must be an x (0 , 4) 3
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2 5 Suppose that the function f x is continuous on the open...

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