Numerical methods to determine the type of roots of an equation

Question

What is the relation between the derivative and the type of roots of a given equation?

I have read once in a book, which I cannot find in PDF (Demidovic - Numerical Analysis) that we can analyze the types of roots (i.e. one real and two complex, or at least one real in $\mathbb{R}^-$, etc.) by looking at the derivative.

For example, take the equation:

$$x^3 - 2x + xe^{-x} = 0$$

What can we say about its roots?

The derivative is

$$3x^2 - 2 + (1-x)e^{-x}$$

I am not talking about the NUMERICAL roots, but just a method which will immediately tell us if that equation will have at least a root in $\mathbb{R}^+$ or every root will be in $\mathbb{R}$ and so on.

I remember it was written in Demidovic's book in the initial pages.

Answer 1

First, look at the limits as $x$ goes to $- \infty$ and $+ \infty$. If answers have different signs, then the number of roots is odds; otherwise it's even. In this case, they have different signs. So there is at least one. To find the number of roots, the derivative will tell you something about this, since there can be at most one root between locations where the derivative is zero.

Answer 2

Observe at first that $x=0$ is a root.

the other roots are solutions of $$f (x)=x^2-2+e^{-x}=0$$

the derivatives are $$f'(x)=2x-e^{-x} $$ and $$f''(x)=2+e^{-x } >0$$ thus $f'(x)=0$ will have at most one root and $f (x)=0$, at most twoo roots.

Graphically, it is clear that $f (x)=0$ has exactly two roots.

Your equation has three roots.