Let $f$ be a polynomial with at least $k$ different roots, prove that $f'$ hast at least $k-1$ different roots

Question

I don't know how to write the proof to this problem:

Let $f$ be a polynomial with at least $k$ different roots, prove that $f'$ has at least $k-1$ different roots.

It's featured among a few exercises made to apply Rolle's theorem, but this one is about a general result and asks for higher degree of formality I guess... I don't know how to write the proof although I have the following intuition:

$f(x)=x^k + g(x)$

$f'(x)=kx^{k-1} + g'(x)$

Any suggestions?

Answer 1

Between any two roots $x_i$ and $x_j$ of $f$, we know that $f$ is continuous on $[x_i, x_j]$ and differentiable on $(x_i, x_j)$ with $f(x_i) = f(x_j) = 0$ so by Rolle's there is a $x_k \in (x_i, x_j)$ such that $f'(x_k) = 0$. Hence $f'$ has at least $k-1 $ roots.

Answer 2

Regards @Gioacchino . I may not give full detail. We can use induction. For $k=2$, then the function $f(x) = (x-r_{1})(x-r_{2})g(x) $ has $$f'(x) = (x-r_{2})g(x) + (x-r_{1})g(x) + (x-r_{1})(x-r_{2})g'(x) $$ The possibilities for $g(x)$ are $g(x) = H(x)$, or $g(x) = (x-r_{1})^{m}H(x)$, or $g(x) = (x-r_{2})^{m}H(x)$ or $g(x) =(x-r_{1})^{m} (x-r_{2})^{n} H(x)$. With these possibilities, you could check that $f'(x)$ will have at least $k-1 = 1$ root.

Now let the statement be true for $k = 2, ... , n$. We would like to prove for $k=n+1$ :

$$ f(x) = (x-r_{1})...(x-r_{n+1})g(x) $$

$$ f'(x) = h(x) + (x-r_{1}) \left[ h'(x) \right] $$

with

$$h(x) = (x-r_{2})...(x-r_{n+1})g(x)$$

Notice that $h(x)$ is a polynomial with at least $n$ different roots, so $h'(x)$ must have at least $n-1$ different roots. We can write $h'(x)$ of the form

$$ h'(x) = (x-r_{m}) G_{m}(x), \:\:\: m = 2,...,n+1 $$

From here it can be seen that $f'(x)$ has at least $k-1 = n$ different roots.

Regards, Arief