Explanation Given: c + c + 2 = 3 The intermediate value theorem states that if a continuous function, f , with an interval [ a , b ] as its domain, takes values f ( a ) and f ( b ) at each end of the interval, then it also takes any value between f ( a ) and f ( b ) at some point within the interval. As a corollary, if a continuous function f ( x ) on the interval [ a , b ] has values of opposite sign inside the interval, then there must be some value x=c in the interval ( a , b ) such that f ( c ) = 0 . To find if c + c + 2 = 3 has a solution, assume the function f ( c ) = c + c + 2 − 3 and show that it is continuous. Then substitute c = 0 , 1 , 2 , 3 and so on to get the corresponding f ( c ) until the corresponding f ( x ) results in one positive and one negative value. A function is continuous on an interval if we can draw the graph from start to finish without ever once picking up our pencil. In other words, a function is continuous if its graph has no holes or breaks in it. The graph of f ( c ) = c + c + 2 − 3 can be drawn as follows, As we can see, the function f ( c ) = c + c + 2 − 3 is continuous on the interval [ 0 , ∞ ) . Calculations: Description f ( c ) = c + c + 2 − 3 Step 1 : Find the f ( c ) values of the corresponding values of c , c = 1 , 2 , 3

3rd Edition

ISBN: 9781464125263

Author: Jon Rogawski, Colin Adams

Publisher: W.H. Freeman & Co

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Chapter 2.8, Problem 7E

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To prove:

The given function has a solution using IVT

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Chapter 2 Solutions

Calculus - Standalone book

Ch. 2.1 - Prob. 1PQ Ch. 2.1 - Prob. 2PQ Ch. 2.1 - Prob. 3PQ Ch. 2.1 - Prob. 4PQ Ch. 2.1 - Prob. 5PQ Ch. 2.1 - Prob. 1E Ch. 2.1 - Prob. 2E Ch. 2.1 - Prob. 3E Ch. 2.1 - Prob. 4E Ch. 2.1 - Prob. 5E

Ch. 2.1 - Prob. 6E Ch. 2.1 - Prob. 7E Ch. 2.1 - Prob. 8E Ch. 2.1 - Prob. 9E Ch. 2.1 - Prob. 10E Ch. 2.1 - Prob. 11E Ch. 2.1 - Prob. 12E Ch. 2.1 - Prob. 13E Ch. 2.1 - Prob. 14E Ch. 2.1 - Prob. 15E Ch. 2.1 - Prob. 16E Ch. 2.1 - Prob. 17E Ch. 2.1 - Prob. 18E Ch. 2.1 - Prob. 19E Ch. 2.1 - Prob. 20E Ch. 2.1 - Prob. 21E Ch. 2.1 - Prob. 22E Ch. 2.1 - Prob. 23E Ch. 2.1 - Prob. 24E Ch. 2.1 - Prob. 25E Ch. 2.1 - Prob. 26E Ch. 2.1 - Prob. 27E Ch. 2.1 - Prob. 28E Ch. 2.1 - Prob. 29E Ch. 2.1 - Prob. 30E Ch. 2.1 - Prob. 31E Ch. 2.1 - Prob. 32E Ch. 2.1 - Prob. 33E Ch. 2.1 - Prob. 34E Ch. 2.1 - Prob. 35E Ch. 2.1 - Prob. 36E Ch. 2.2 - Prob. 1PQ Ch. 2.2 - Prob. 2PQ Ch. 2.2 - Prob. 3PQ Ch. 2.2 - Prob. 4PQ Ch. 2.2 - Prob. 5PQ Ch. 2.2 - Prob. 6PQ Ch. 2.2 - Prob. 7PQ Ch. 2.2 - Prob. 1E Ch. 2.2 - Prob. 2E Ch. 2.2 - Prob. 3E Ch. 2.2 - Prob. 4E Ch. 2.2 - Prob. 5E Ch. 2.2 - Prob. 6E Ch. 2.2 - Prob. 7E Ch. 2.2 - Prob. 8E Ch. 2.2 - Prob. 9E Ch. 2.2 - Prob. 10E Ch. 2.2 - Prob. 11E Ch. 2.2 - Prob. 12E Ch. 2.2 - Prob. 13E Ch. 2.2 - 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Prob. 56E Ch. 2.5 - Prob. 57E Ch. 2.5 - Prob. 58E Ch. 2.5 - Prob. 59E Ch. 2.6 - Prob. 1PQ Ch. 2.6 - Prob. 2PQ Ch. 2.6 - Prob. 3PQ Ch. 2.6 - Prob. 1E Ch. 2.6 - Prob. 2E Ch. 2.6 - Prob. 3E Ch. 2.6 - Prob. 4E Ch. 2.6 - Prob. 5E Ch. 2.6 - Prob. 6E Ch. 2.6 - Prob. 7E Ch. 2.6 - Prob. 8E Ch. 2.6 - Prob. 9E Ch. 2.6 - Prob. 10E Ch. 2.6 - Prob. 11E Ch. 2.6 - Prob. 12E Ch. 2.6 - Prob. 13E Ch. 2.6 - Prob. 14E Ch. 2.6 - Prob. 15E Ch. 2.6 - Prob. 16E Ch. 2.6 - Prob. 17E Ch. 2.6 - Prob. 18E Ch. 2.6 - Prob. 19E Ch. 2.6 - Prob. 20E Ch. 2.6 - Prob. 21E Ch. 2.6 - Prob. 22E Ch. 2.6 - Prob. 23E Ch. 2.6 - Prob. 24E Ch. 2.6 - Prob. 25E Ch. 2.6 - Prob. 26E Ch. 2.6 - Prob. 27E Ch. 2.6 - Prob. 28E Ch. 2.6 - Prob. 29E Ch. 2.6 - Prob. 30E Ch. 2.6 - Prob. 31E Ch. 2.6 - Prob. 32E Ch. 2.6 - Prob. 33E Ch. 2.6 - Prob. 34E Ch. 2.6 - Prob. 35E Ch. 2.6 - Prob. 36E Ch. 2.6 - Prob. 37E Ch. 2.6 - Prob. 38E Ch. 2.6 - Prob. 39E Ch. 2.6 - Prob. 40E Ch. 2.6 - Prob. 41E Ch. 2.6 - Prob. 42E Ch. 2.6 - Prob. 43E Ch. 2.6 - Prob. 44E Ch. 2.6 - 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Prob. 31E Ch. 2.7 - Prob. 32E Ch. 2.7 - Prob. 33E Ch. 2.7 - Prob. 34E Ch. 2.7 - Prob. 35E Ch. 2.7 - Prob. 36E Ch. 2.7 - Prob. 37E Ch. 2.7 - Prob. 38E Ch. 2.7 - Prob. 39E Ch. 2.7 - Prob. 40E Ch. 2.7 - Prob. 41E Ch. 2.7 - Prob. 42E Ch. 2.7 - Prob. 43E Ch. 2.7 - Prob. 44E Ch. 2.7 - Prob. 45E Ch. 2.8 - Prob. 1PQ Ch. 2.8 - Prob. 2PQ Ch. 2.8 - Prob. 3PQ Ch. 2.8 - Prob. 4PQ Ch. 2.8 - Prob. 5PQ Ch. 2.8 - Prob. 1E Ch. 2.8 - Prob. 2E Ch. 2.8 - Prob. 3E Ch. 2.8 - Prob. 4E Ch. 2.8 - Prob. 5E Ch. 2.8 - Prob. 6E Ch. 2.8 - Prob. 7E Ch. 2.8 - Prob. 8E Ch. 2.8 - Prob. 9E Ch. 2.8 - Prob. 10E Ch. 2.8 - Prob. 11E Ch. 2.8 - Prob. 12E Ch. 2.8 - Prob. 13E Ch. 2.8 - Prob. 14E Ch. 2.8 - Prob. 15E Ch. 2.8 - Prob. 16E Ch. 2.8 - Prob. 17E Ch. 2.8 - Prob. 18E Ch. 2.8 - Prob. 19E Ch. 2.8 - Prob. 20E Ch. 2.8 - Prob. 21E Ch. 2.8 - Prob. 22E Ch. 2.8 - Prob. 23E Ch. 2.8 - Prob. 24E Ch. 2.8 - Prob. 25E Ch. 2.8 - Prob. 26E Ch. 2.8 - Prob. 27E Ch. 2.8 - Prob. 28E Ch. 2.8 - Prob. 29E Ch. 2.8 - Prob. 30E Ch. 2.8 - Prob. 31E Ch. 2.8 - 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To prove: The given function has a solution using IVT

Solution Summary: The author proves that the given function has a solution using IVT. The intermediate value theorem states that if the continuous function f(c)=sqrt