Pyruvic Acid
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If there is a change in sign in f(x) within a given interval, there must be either no roots or an even number of roots in that interval. Is this correct?

What about for f(x) = (2-x)2? There is no change of sign in the interval [0,3], yet there are an odd number of roots within this interval, no? Does the fact that the root is repeated have something to do with it? Do we count the change from f(x) = +ve to f(x) = 0 (and vice versa) as a change in sign?
Last edited by Pyruvic Acid; 4 weeks ago
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RDKGames
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(Original post by Pyruvic Acid)
If there is a change in sign in f(x) within a given interval, there must be either no roots or an even number of roots in that interval. Is this correct?

What about for f(x) = (2-x)2? There is no change of sign in the interval [0,3], yet there are an odd number of roots within this interval, no? Does the fact that the root is repeated have something to do with it? Do we count the change from f(x) = +ve to f(x) = 0 (and vice versa) as a change in sign?
For a continuous f(x) the intermediate value theorem says that if there is a change in sign of f between two points, there is AT LEAST one root between the two points.

Anyway, a change in sign guarantees at least one root,
BUT the converse is not true; no change in sign does not guarantee that there are no roots. As per your example: (2-x)^2 has one root in interval [0,3] yet no change of sign. A change of sign is when you go from <0 to >0 or vice versa.
Last edited by RDKGames; 4 weeks ago
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Pyruvic Acid
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(Original post by RDKGames)
For a continuous f(x) the intermediate value theorem says that if there is a change in sign of f between two points, there is AT LEAST one root between the two points.

Anyway, a change in sign guarantees at least one root,
BUT the converse is not true; no change in sign does not guarantee that there are no roots. As per your example: (2-x)^2 has one root in interval [0,3] yet no change of sign. A change of sign is when you go from <0 to >0 or vice versa.
Thanks!
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ghostwalker
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(Original post by Pyruvic Acid)
If there is a change in sign in f(x) within a given interval, there must be either no roots or an even number of roots in that interval. Is this correct?

What about for f(x) = (2-x)2? There is no change of sign in the interval [0,3], yet there are an odd number of roots within this interval, no? Does the fact that the root is repeated have something to do with it? Do we count the change from f(x) = +ve to f(x) = 0 (and vice versa) as a change in sign?
4th attempt, having realised my previous ones were not quite correct and needed more qualifying than was useful.

At a given root of a polynomial, you will get a sign change if, and only if, the exponent of the corresponding factor is odd.

E.g in your example, there is a root at x=2. The corresponding factor is (x-2)^2. The exponent is 2 (i.e. even) so no sign change.
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Pyruvic Acid
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(Original post by ghostwalker)
4th attempt, having realised my previous ones were not quite correct and needed more qualifying than was useful.

At a given root of a polynomial, you will get a sign change if, and only if, the exponent of the corresponding factor is odd.

E.g in your example, there is a root at x=2. The corresponding factor is (x-2)^2. The exponent is 2 (i.e. even) so no sign change.
What about for x2+7x+12 in the interval [-4,-2]? Although the exponent of corresponding factor is even, there is a change in sign, no?
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ghostwalker
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(Original post by Pyruvic Acid)
What about for x2+7x+12 in the interval [-4,-2]? Although the exponent of corresponding factor is even, there is a change in sign, no?
What is it in factor format? (x-a)(x-b)

How do you know the corresponding factor is even?
Last edited by ghostwalker; 4 weeks ago
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Pyruvic Acid
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(Original post by ghostwalker)
What is it in factor format? (x-a)(x-b)

How do you know the corresponding factor is even?
Ahh, I initially misunderstood what corresponding factor meant. Thanks for your help!
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