Proof Let a 0 Set p x x 2 a on 0 b where b 2 a For example let b a 1 Then p 0 a

Proof let a 0 set p x x 2 a on 0 b where b 2 a for

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Proof: Let a > 0. Set p ( x ) = x 2 - a on [0 , b ] where b 2 > a . For example, let b = a + 1. Then, p (0) = - a < 0 and p ( b ) = b 2 - a > 0. Therefore, by the Intermediate-Value Theorem, there exists at least one number c (0 , b ) such that p ( c ) = 0. That is, c 2 - a = 0, which implies c = a . 5. Fix a positive number P . Let R be the set of all rectangles with perimeter P = 4. Prove that there is a member of R that has maximum area. Proof: Let R be a member of R , and let R have length x and width y . Since the perimeter is 4, 2 x + 2 y = 4 so y = 2 - x . The area of R is A ( x ) = x (2 - x ) = 2 x - x 2 , 0 x 2 (since x and y are each non-negative). Now A is a continuous function on the compact set D = [0 , 2]. Therefore A has a maximum value (and a minimum value which is 0). That is, there is a number c [0 , 2] such that A ( x ) A ( c ) for all x [0 , 2].
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6. Let f : [ a, b ] R and g : [ a, b ] R be continuous functions. Prove that if f ( a ) < g ( a ) and g ( b ) < f ( b ), then there is at least one number c ( a, b ) such that f ( c ) = g ( c ). Proof: Set h ( x ) = f ( x ) - g ( x ). Then h : [ a, b ] R is continuous, h ( a ) = f ( a ) - g ( a ) < 0 and h ( b ) = f ( b ) - g ( b ) > 0. Therefore, by the Intermediate- Value Theorem, there is a number c ( a, b ) such that h ( c ) = f ( c ) - g ( c ) = 0 which implies f ( c ) = g ( c ). 7. Let f : D R . Prove that if f is bounded on a neighborhood of each x D and D is compact, then f is bounded on D . Proof: For each x D there is a neighborhood N ( x ) of x and a number M x such that | f ( x ) | ≤ M x for all x N ( x ). The collection of neighborhoods N ( x ) , x D is an open cover of D . Since D is compact, this open cover has a finite subcover, N ( x 1 ) , N ( x 2 ) , · · · , N ( x k ). Associated with each neighborhood N ( x p ) , 1 p k there is a number M x p . Let M = max M x 1 , M x 2 , · · · , M x p . Now, choose any point z D . Then z N ( x p ) for some p ∈ { 1 , 2 , · · · , k } . This implies that, | f ( z ) | ≤ M x p M . Therefore we can conclude that | f ( z ) | ≤ M for all z D which means f is bounded on D .
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8. Use the definition of the derivative to find the derivative of f ( x ) = 4 + x at x = 5. METHOD 1: lim x 5 f ( x ) - f (5) x - 5 = lim x 5 4 + x - 3 x - 5 = lim x 5 4 + x - 3 x - 5 · 4 + x + 3 4 + x + 3 = lim x 5 x - 5 ( x - 5)( 4 + x + 3) = lim x 5 1 4 + x + 3 = 1 6 Therefore, f 0 (5) = 1 6 . METHOD 2: Alternatively, in terms of h , we have lim h 0 f (5 + h ) - f (5) h = lim h 0 p 4 + (5 + h ) - 3 h = lim h 0 9 + h - 3 h · 9 + h + 3 9 + h + 3 = lim h 0 h h ( 9 + h + 3) = lim h 0 1 9 + h + 3 = 1 6 Therefore, f 0 (5) = 1 6 .
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9. Use the definition of the derivative to find the derivative of f ( x ) = 1 2 x + 1 . METHOD 1: For any number c 6 = - 1 / 2, we have lim x c f ( x ) - f ( c ) x - c = lim x c 1 2 x + 1 - 1 2 c + 1 x - c = lim x c (2 c + 1) - (2 x + 1) (2 x + 1)(2 c + 1) x - c = lim x c (2 c + 1) - (2 x + 1) ( x - c )(2 x + 1)(2 c + 1) = lim x c - 2 (2 x + 1)(2 c + 1) = - 2 (2 c + 1) 2 Therefore, f 0 ( x ) = - 2 (2 x + 1) 2 , for x 6 = - 1 2 . METHOD 2: For any number c 6 = - 1 / 2, we have lim h 0 f ( c + h ) - f ( c ) h = lim h 0 1 2( c + h ) + 1 - 1 2 c + 1 h = lim h 0 (2 c + 1) - (2 c + 2 h + 1) (2 c + 2 h + 1)(2 c + 1) h = lim h 0 (2 c + 1) - (2 c + 2 h + 1) h (2 c + 2 h + 1)(2 c + 1) = lim h 0 - 2 h h (2 c + 2 h + 1)(2 c + 1) = lim h 0 - 2 (2 c + 2 h + 1)(2 c + 1) = - 2 (2 c + 1) 2 Therefore, f 0 ( x ) = - 2 (2 x + 1) 2 , for x 6 = - 1 2 .
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10. Set f ( x ) = ( x 3 + 3 x + 2 , x 1 3 x 2 + 3 , x > 1 .
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