# Root Question

1. Nov 5, 2007

### grapeape

1. The problem statement, all variables and given/known data
Show that f(x) = x^4 + 4x + c = 0 has at most 2 roots.

2. Relevant equations

3. The attempt at a solution
I'm not really sure how to approach this problem, I think I have to use the IMVT / Rolle's Theorem / MVT.

Any help to even get me started would be greatly appreciated!

2. Nov 5, 2007

### HallsofIvy

Staff Emeritus
Rolle's theorem sounds good! What is the derivative of x4+ 4x+ c? Suppose there were in fact three differerent roots. Apply Rolle's theorem on the two different intervals.

Last edited: Nov 6, 2007
3. Nov 5, 2007

### andytoh

It's derivative has only one root, and so the f(x) has only one critical point

4. Nov 5, 2007

### grapeape

the derivative is 4x^3 + 4, which has a zero at -1. But how do I eliminate the possibility of other zeroes? do I have to take the double deriv? and also how do I prove the f(x) has only 2 zeroes?

5. Nov 5, 2007

### andytoh

-1 is the only critical point of f(x). Draw a picture to convince yourself that f(x) has at most two roots.

6. Nov 5, 2007

### grapeape

So, since there is only 1 critical point, how do we know that there aren't more? asides from using a graphing calculator. By looking at the first derivative and finding 1 root, does that automatically mean that there are two roots? I'm a little confused on how Rolle's theorem ties into this. Am I supposed to look at the second derivative? So far, I've written that since there is only one critical point, it means that the graph of f(x) only changes from decr. to incr. once, so there can only be two zeros? is that right?

7. Nov 5, 2007

### andytoh

There's only one real solution to x^3 = -1, the other two solutions are complex.

8. Nov 5, 2007

### grapeape

alright, thanks. I am pretty sure I got it now. I used IMVT to prove that between -2, and -1 there is another 0, and since there is only 1 critical number it is the only other 0 asides from (0,0). Seems right, thanks for the help Andy and Halls!