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Unformatted text preview: Wharton Math Camp Problem Set 3 – Solutions 1. A function f : [ a,b ] ⊂ R is continuous if for any two points x 1 ,x 2 ∈ [ a,b ] such that x 1 < x 2 and for any real number c lying between f ( x 1 ) and f ( x 2 ), there is x ∈ ( x 1 ,x 2 ) such that f ( x ) = c . Prove or give a counterexample. (Hint: use Example 1.51 in Sundaram) Solution . Following Sundaram, Example 1.51, let f ( x ) = if x = 0 x 2 cos(1 /x ) if x 6 = 0. Notice that f is not continuous at 0 because lim x → = ∞ , but f (0) = 0. We want to show that even if not continuous, this function has the Intermediate Value Property. Given some a < 0, we need to show that for every c ∈ ( f ( a ) , 0), there exists a x ∈ ( a, 0) such that f ( x ) = c . The proof that such an x exists relies on the fact that f is continuous everywhere, except at zero. Observe that we may choose some y in ( a, 0) so that f ( y ) = 0. To see this, note that the sign of cos( a + π ) equals the sign of cos( a ). Hence, we just need to show that there exists some w in ( a, 0) such that 1 a + w = 1 a + π . It turns out that w = πa 2 1+ πa suffices for a such that  a  < 1. Now, we may use the Intermediate Value Theorem to find some x ∈ ( a,y ) so that f ( x ) = c . This suffices since ( a,y ) ⊂ ( a, 0). Notice that the function f given in Sundaram’s example is not a coun terexample for this problem, since it is differentiable (hence continuous) everywhere. The first derivative though, is a good counterexample. 2. Sundaram, Chapter 2, Exercise 1. Solution . Sundaram, Theorem 2.5 is not valid if ϕ : R → R is non decreasing. To see this, consider the following counterexample. Let f : [ 1 , 1] → R be defined as f ( x ) = x 1 + 1 . Then x * = 0 is a maximizer for f . Choose now ϕ ( x ) = 0 for all x ∈ [ 1 , 1], and notice that ϕ is non decreasing. Then, ϕ ( f ( x )) = 0 for all x ∈ [ 1 , 1]. Every x ∈ [ 1 , 1] is a maximizer for ϕ ◦ f , but only x * = 0 is a maximizer for f . 2 3. Sundaram, Chapter 2, Exercise 6. Solution . Let f : [0 , 1] → R be defined as f ( x ) = 1 2 , if x = 0 x, if x ∈ (0 , 1) 1 2 , if x = 1 . Then, f ([0 , 1]) = (0 , 1) which is open. 3 4. Sundaram, Chapter 2, Exercise 8. Solution . The statement is false. Consider the following counterex ample. Let f ( x ) = 3 x and let g be defined as g ( x ) = x, if x ∈ [0 , 1) 2 , if x = 1 , and notice that f is strictly increasing, and g is strictly decreasing. Then, ( f + g )( x ) = 2 x, if x ∈ [0 , 1) +1 , if x = 1 , which does not achieve a maximum since ( f + g )([0 , 1]) = [0 , 2). 4 5. Sundaram, Chapter 3, Exercise 3. Solution . Let f : D ⊂ R → R . Since f in nondecreasing, we have that f ( x ) ≤ f ( y ) for all x < y ∈ D . A maximum for f is the point x * such that f ( x * ) ≥ f ( y ) for all y ∈ D ....
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 Spring '02
 Ms.Kirson
 Math, Optimization, y1, negative definite

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