Show that the equation has at most one root

[Painguy]

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?

 

[LCKurtz]

Hint: Look at the derivative.

 

[HallsofIvy]

Blast you, LCKurtz!

I was going to say that!

 

[Painguy]

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?

 

[LCKurtz]

Blast you, LCKurtz!

I was going to say that!

Heh heh. Go ahead and say it. 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".

 

[LCKurtz]

So i took the derivative and got:
f'(x)=5X^4 -6

The derivative of 10x is -6?

 

[Painguy]

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

 

[LCKurtz]

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...?

 

[Painguy]

So the derivative is..., and therefore...?

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

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

therefore there is only 1 root?

 

[HallsofIvy]

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?

 

[LCKurtz]

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...?

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

The derivative of -6 is not -6.

 

[Painguy]

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

 

[Mark44]

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.