Explanation Given: x 6 − 8 x 4 + 10 x 2 − 1 = 0 Key concepts applied: 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 x 6 − 8 x 4 + 10 x 2 − 1 = 0 has a solution in a given interval, assume the function f ( x ) = x 6 − 8 x 4 + 10 x 2 − 1 and show that it is continuous. Then substitute x = − 3 , − 2 , − 1 , 0 , 1 , 2 , 3 and so on to get the corresponding f ( x ) until the corresponding function results in one positive and one negative value or vice versa. The function is then said to have a solution in that interval. Since we have to check if the function has at least six distinct solutions, we test the values until we get six such intervals. 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 ( x ) = x 6 − 8 x 4 + 10 x 2 − 1 can be drawn as follows, As we can see, the function f ( x ) = x 6 − 8 x 4 + 10 x 2 − 1 is continuous on the interval [ − 3 , 3 ] . Calculations: Description f ( x ) = x 6 − 8 x 4 + 10 x 2 − 1 Step 1 : Find the f ( c ) values of the corresponding values of c , c = − 3 , − 2 , − 1 , 0 , 1 , 2 , a n d 3 f ( − 3 ) = ( − 3 ) 6 − 8 ( − 3 ) 4 + 10 ( − 3 ) 2 − 1 = 729 − 648 + 90 − 1 = 170 ( p o s i t i v e ) f ( − 2 ) = ( − 2 ) 6 − 8 ( − 2 ) 4 + 10 ( − 2 ) 2 − 1 = 64 − 128 + 40 − 1 = − 25

4th Edition

ISBN: 9781319323394

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 2.8, Problem 17E

To determine

To prove:

The given function has at least six distinct solutions using the Intermediate Value Theorem

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Chapter 2.8, Problem 16E

Chapter 2.8, Problem 18E

Chapter 2 Solutions

CALCULUS W/SAPLING ACCESS >IC<

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

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To prove: The given function has at least six distinct solutions using the Intermediate Value Theorem

Solution Summary: The author proves that the given function has at least six distinct solutions using the Intermediate Value Theorem.