Show equation has Exactly one real root


by jkeatin
Tags: equation, real, root
jkeatin
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#1
Jul6-08, 09:37 PM
P: 66
1. The problem statement, all variables and given/known data

show that the equation 1 + 2x + x^3 + 4x^5 = 0 has exactly one real root

2. Relevant equations



3. The attempt at a solution

i dont know what to use to find out that it has exactly one real root. i know you use the intermediate value theorem for roots but what do you do to show it has exactly one?
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durt
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#2
Jul6-08, 09:55 PM
P: 152
First show that it has at least one real root (you can use the intermediate value theorem, as you noted). Then consider its derivative. What does it tell you about this function?
jkeatin
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#3
Jul6-08, 10:26 PM
P: 66
i used f(1) and f(-1) to prove the intermediate value theorem. the derivative of the function is 2+3x^2+20x^4 which i guess shows that the function is always positive?

Dick
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#4
Jul6-08, 11:38 PM
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Show equation has Exactly one real root


Quote Quote by jkeatin View Post
i used f(1) and f(-1) to prove the intermediate value theorem. the derivative of the function is 2+3x^2+20x^4 which i guess shows that the function is always positive?
It doesn't show the function is always positive. It shows that its always increasing. If it has a root, what does this tell you about the number of roots?
re182
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#5
Nov20-08, 07:30 PM
P: 1
Take two numbers, say -1 and 0. Then f(-1) = -6 < 0 and f(0) = 1 > 0 (Letting f(x) = 1 + 2x + x^3 + 4x^5.)

By the Intermediate Value Theorem there exists a number "c" between -1 and 0 such that f(c) = 0. So the equation has a real root. Then suppose there are two roots a and b.

Then f(a) = f(b) = 0 and by Rolle's Theorem f'(r) = 0 for some r in the set of numbers (a,b). But f'(x) = 2 + 3x^2 + 20x^4 > 0 for all x, so it is impossible to have f'(r) = 0 for some c.

So there is exactly one root.
cmw1229
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#6
Apr3-09, 12:52 AM
P: 1
Quote Quote by re182 View Post
Take two numbers, say -1 and 0. Then f(-1) = -6 < 0 and f(0) = 1 > 0 (Letting f(x) = 1 + 2x + x^3 + 4x^5.)

By the Intermediate Value Theorem there exists a number "c" between -1 and 0 such that f(c) = 0. So the equation has a real root. Then suppose there are two roots a and b.

Then f(a) = f(b) = 0 and by Rolle's Theorem f'(r) = 0 for some r in the set of numbers (a,b). But f'(x) = 2 + 3x^2 + 20x^4 > 0 for all x, so it is impossible to have f'(r) = 0 for some c.

So there is exactly one root.
Can anyone explain this is simpler terms. I grasp it for the most part but I don't see the relationship between the derivative being positive (Increasing) and having only one root on that interval.
FedEx
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#7
Apr3-09, 05:27 AM
P: 336
Quote Quote by Dick View Post
It doesn't show the function is always positive. It shows that its always increasing. If it has a root, what does this tell you about the number of roots?
Perfect. Forget everything. Just look int tothe logic. Always increasing. Hence it either intersects the x axis once or not at all. But it is evident that the function exists in negative y and we can alos see that it exists in +ve y alos. So it intersects the x axis once. Cause to intersect the x axis more than once the function has to decrease. Bur we know that that doesnot happen


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