use rolle's theorem to show that x^3+2x+c=0, c a constant cannot have more than one real root?

Question

please give me full answer..tq!

Answer 1

Rolle's theorem (paraphrased) says that if a continuous differentiable function f takes on the same value at two different points, then we can find a point, c, between these points such that f'(c) = 0

Ok, let's look at the derivative of this equation...
3x^2 + 2 = 0....
so 3x^2 = -2
x^2 = (-3/2)
x = +/- i sqrt (3/2)
That means that this equation has no real critical values, in other words, no real numbers where it takes on a minimum or maximum value. The function appears to be 1:1 which means it would cross the x axis in at most one point. So there are no two points a and b such that f(a) = f(b), so Rolle's theorem does not apply.
It appears you're assuming the converse of Rolle's theorem as well, which you can't assume. Perhaps you could attack this question from the point of view of the contrapositive. This isn't a complete answer, sorry. But it's a start, and I've got some things to do this morning, so maybe you can figure it out from here or someone else can help you.

Answer 2

Yeh, I am also wondering why Rolle's Theorem applies, since you need two values of x that give the same answer to use that theorem.

As others have said, the derivative of your function is 3x^2 + 2, which is always greater than zero, i.e. the function is always increasing. Clearly, if a continuous function is always increasing, it can only cross the x-axis once - it would have to start decreasing to cross it again.

Answer 3

Let f(x) = x^3+2x+c.
f'(x) = 3x^2+2 is always greater than zero. If you can find at least two points such that f(c1) = f(c2) = 0, then by Rolle's theorem, you have a point c between c1 and c2 such that f'(c) = 0. This is contradictary to f'(x) is always greater than zero.Therefore, you can not find two points c1 and c2 such that f(c1) = f(c2) = 0.