# Proving roots using mean value theorem

1. Mar 21, 2006

### Jeff Ford

Prove $$x^4 + 4x + c = 0$$ has at most two real roots

My thinking is that to prove this I would assume that it has three real roots and look for a contradiction.

So I set $$f(x) = x^4 + 4x + c$$ and assume three real roots $$x_1, x_2, x_3$$ such that $$f(x_1) = f(x_2) = f(x_3) = 0$$

By MVT I know that there must exist $$c_1$$ on the interval $$(x_1, x_2)$$ such that $$f'(c_1) = 0$$ and $$c_2$$ on the interval $$(x_2, x_3)$$ such that $$f'(c_2) = 0$$ and $$c_3$$ on the interval $$(x_1, x_3)$$ such that $$f'(c_3) = 0$$

Now I'm a little confused. Do I try to find three values of c that will satisfy $$f'(x) = 4x^3 + 4 = 0$$? Only the value x = -1 would make this true, so is that my contradiction? That MVT predicts three values and only one exists?

Thanks
Jeff

Last edited: Mar 21, 2006
2. Mar 21, 2006

### arildno

Hint:
How many STATIONARY points does the function have?

3. Mar 21, 2006

### Jeff Ford

Only one, at (-1, 5+c). Since the derivative is negative to the left of that point and positive to the right, it must be a minimum of the function.

Last edited: Mar 21, 2006
4. Mar 21, 2006

### Galileo

I assume you orden x1 < x2 < x3. It could be the case that f'(c3) equals f'(c1) or f'(c2), so according to the MVT you'll get at least 2 points where f' is zero. You've shown there's only one, so the assumption of three distinct roots is false. Not sure why you want to find particular values of c to make f' zero. The value of c is irrelevant to the problem.

5. Mar 21, 2006

### Jeff Ford

Thanks for the clarification