Question about polynomials and its derivatives

Question

Show that if the polynomial that has all real zeros but without multiple roots, has the properties that all its derivatives till the derivative of order $n-1$ have also real zeros.

Can somebody help me with this problem? I thought that if we take $f(x)$ that satisfies the given conditions we can write it as $f(x)=c(x-x_1)(x-x_2) \cdots (x-x_n)$

Where $x_1,x_2, \cdots ,x_n$ are the real zeros of $f(x)$

Then we have $$f'(x)=\frac{cf(x)}{x-x_1}+\frac{cf(x)}{x-x_2}+ \cdots +\frac{cf(x)}{x-x_n}$$

Where $f'(x_i) \ne0$ for $i \in {1,2, \cdots , n }$

But I don't know how to prove that $f'(x)$ has real zeros. Or should I approach it differently?

I'm really sorry for any mistakes in my English. It's not my native language.

Answer 1

The derivative of a smooth (all its derivatives exist) function between two zeros of degree $1$ must change sign. Why?

Apply this argument inductively (the second derivative is the derivative of the first derivative, and so on).

Answer 2

Since $f$ is continuous, its first derivative must reach zero at least once inside the interval $[x_{I}, x_{I+1}]$. This accounts for all its zeros since there's exactly $n-1$ of them. An induction argument completes the proof for higher order derivatives.

Answer 3

A polynomial with no multiple roots alternates in sign and has extrema between two successive roots. Hence its derivative has at least as many roots, minus one.

On the other hand, it cannot have more, as the number of roots cannot exceed the degree.