# Show that the equation has at most one root

1. Oct 15, 2012

### Painguy

1. The problem statement, all variables and given/known data
Show that x^5 +10x +c=0 has at most one real root on the interval [-1, 1]

2. Relevant equations

3. 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?

2. Oct 15, 2012

### LCKurtz

Hint: Look at the derivative.

3. Oct 15, 2012

### HallsofIvy

Blast you, LCKurtz!

I was going to say that!

4. Oct 15, 2012

### Painguy

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. Oct 15, 2012

### LCKurtz

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".

6. Oct 15, 2012

### LCKurtz

The derivative of 10x is -6??????

7. Oct 15, 2012

### Painguy

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

x^5 -6 +c =0

8. Oct 16, 2012

### LCKurtz

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

9. Oct 16, 2012

### Painguy

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

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

therefore there is only 1 root?

10. Oct 16, 2012

### 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?

11. Oct 16, 2012

### LCKurtz

The derivative of -6 is not -6.

12. Oct 16, 2012

### Painguy

Wow whats wrong with me I meant to say

x^5 -6x +c =0

Please forgive my stupidity. :P

13. Oct 16, 2012

### Staff: Mentor

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.