Proving Roots of f(x) with Odd n ≤ 3

[Neoma]

Let n \ \epsilon \ \mathbb{N}, \ n \geq 2 and p, \ q \ \epsilon \ \mathbb{R}. Consider f: \ \mathbb{R} \ \rightarrow \ \mathbb{R} defined by f(x)=x^{n}+px+q.

Suppose n is odd, prove that f has at least one and at most three real roots.

I thought about the intermediate value theorem for proving that f has one root, but then you need one x where f is negative and another one where it's positive and it's impossible to expres this x in terms of n, p and q.

To prove that f has at most three real roots, I thought about finding the local extrema (where f'(x)=0) and examining each of the possible combinations of positions of them. However, then I'm kinda facing the same problem. I'm sure there has to be some more elegant way.

 

[PhilG]

Let n \ \epsilon \ \mathbb{N}, \ n \geq 2 and p, \ q \ \epsilon \ \mathbb{R}. Consider f: \ \mathbb{R} \ \rightarrow \ \mathbb{R} defined by f(x)=x^{n}+px+q.

Suppose n is odd, prove that f has at least one and at most three real roots.

I thought about the intermediate value theorem for proving that f has one root, but then you need one x where f is negative and another one where it's positive and it's impossible to expres this x in terms of n, p and q.

You have the right idea. For large absolute values of x, what is the dominant term of f(x)? In other words, what is the limit of f(x) as x goes to infinity, minus infinity?

To prove that f has at most three real roots, I thought about finding the local extrema (where f'(x)=0) and examining each of the possible combinations of positions of them. However, then I'm kinda facing the same problem. I'm sure there has to be some more elegant way.

Again, you are using the right approach by considering the equation f'(x) = 0. Consider separately the cases when p is positive, negative, and zero. In each case, how many real roots can f'(x) = 0 have?

 

[Neoma]

Thanks, I solved it.