Show that the equation has at most one root

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In summary: Therefore it is not surprising that you get two different values of c.Hint: Look at the derivative.In summary, the conversation involves finding the number of real roots of the equation x^5 + 10x + c = 0 on the interval [-1, 1] and suggests using the derivative to determine this. The solution is that there can be at most one real root on this interval, and the value of c needs to be chosen carefully in order for this to be the case.
  • #1
Painguy
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Homework Statement


Show that x^5 +10x +c=0 has at most one real root on the interval [-1, 1]

Homework Equations



The Attempt at a Solution



I first try to find an x value inorder to find c so i take the end points of the interval and solve for for c

-1 +6=-c
-5=c

1-6=-c
-5=cI get two c's which isn't right so I'm guessing depending on which c i choose there is only 1 root?
 
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  • #2
Hint: Look at the derivative.
 
  • #3
Blast you, LCKurtz!

I was going to say that!
 
  • #4
LCKurtz said:
Hint: Look at the derivative.
So i took the derivative and got:
f'(x)=5X^4 -6

So i plug in the end points and the outputs seem to be the same which means that f'(x)<0 on [-1,1] meaning that it can only cross the x-axis once along that interval, correct?
 
  • #5
HallsofIvy said:
Blast you, LCKurtz!

I was going to say that!

Heh heh. Go ahead and say it. :smile: It's OK to repeat, like Erdos did when he gave a "silver spoon" argument: "Chebyshev said it, and I say it again, There is always a prime between n and 2n".
 
  • #6
Painguy said:
So i took the derivative and got:
f'(x)=5X^4 -6

The derivative of 10x is -6?
 
  • #7
LCKurtz said:
The derivative of 10x is -6?

woops it seems i wrote the wrong equation in the original post. The equation in question is

x^5 -6 +c =0

sorry about that
 
  • #8
Painguy said:
woops it seems i wrote the wrong equation in the original post. The equation in question is

x^5 -6 +c =0

sorry about that

So the derivative is..., and therefore...?
 
  • #9
LCKurtz said:
So the derivative is..., and therefore...?
f'(x)=5X^4 -6

f'(x)<0 on [-1,1]

therefore there is only 1 root?
 
  • #10
Between any two roots of a polynomial, there is a root of its derivative. Since f' has no root in [-1, 1], f can has no more than one root in that interval.

Now, what can you say about c so that there exists exactly one root in that interval?
 
  • #11
Painguy said:
woops it seems i wrote the wrong equation in the original post. The equation in question is

x^5 -6 +c =0

sorry about that

LCKurtz said:
So the derivative is..., and therefore...?

Painguy said:
f'(x)=5X^4 -6

The derivative of -6 is not -6.
 
  • #12
LCKurtz said:
The derivative of -6 is not -6.

Wow what's wrong with me I meant to say

x^5 -6x +c =0

Please forgive my stupidity. :P
 
  • #13
Painguy said:

Homework Statement


Show that x^5 +10x +c=0 has at most one real root on the interval [-1, 1]


Homework Equations



The Attempt at a Solution



I first try to find an x value inorder to find c so i take the end points of the interval and solve for for c

-1 +6=-c
-5=c

1-6=-c
-5=c


I get two c's which isn't right so I'm guessing depending on which c i choose there is only 1 root?

Aside from the fact that this isn't the right approach, you are making two tacit assumptions.
1. When you substitute x = -1, you are assuming that the root is at x = -1, and looking for the value of c that makes that happen.
2. When you substitute x = 1, you are now assuming that the root is at x = +1, and looking for the almost certainly different value of c that makes that happen.

It would be extremely unusual for the same value of c to work in both cases.
 

FAQ: Show that the equation has at most one root

What does it mean for an equation to have at most one root?

When an equation has at most one root, it means that there is at most one value that satisfies the equation and makes it true. This means that the equation only has one solution.

How can you prove that an equation has at most one root?

To prove that an equation has at most one root, you can use the intermediate value theorem. This theorem states that if a continuous function has different signs at two points, then it must have at least one root in between those two points. If the function only changes signs once, then it can have at most one root.

Can an equation have more than one root if it has at most one root?

No, if an equation has at most one root, it means that there can only be one value that satisfies the equation. Therefore, it cannot have more than one root.

Are there any exceptions to the rule that an equation with at most one root can only have one solution?

Yes, there are some exceptions to this rule. For example, if the equation is a quadratic with a double root, it would technically have two roots but they would be the same value. In general, these exceptions occur when there is a repeated root or a complex root.

How can knowing that an equation has at most one root be useful in solving mathematical problems?

Knowing that an equation has at most one root can be useful in many ways. For example, it can help us narrow down the possible solutions when solving equations. It can also help us determine the maximum or minimum values of a function by finding the single root of its derivative. Additionally, it can be applied in various real-life situations where we need to find the single solution to a problem.

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