Let $f(x)=(x^2-1)^n$. Prove that for $r=0,1, ... ,n$, $f^{(r)}(x)$ is a polynomial whose value is $0$ at no fewer than $r$ distinct points on $(-1,1)$. In other words, prove that $f^{(n)}(x)$.
I know that I am supposed to perform an induction. Moreover, it is apparent that if we expand the function $f$ that we have binomial expansions at each order of the derivative.
I know that this question has been asked, but I need help showing that the $r-th$ derivative is a polynomial, i.e. that $f$ is differentiable at every order. I can figure out the rest from their.
This just follows by applying Rolle's theorem and induction on $r$.
Hints:
$f(x)$ has a stationary point in $x=0$ since $f(x)=f(-x)$, hence $f'(0)=0$.
$f'(x)$ is zero in $x=\pm 1$, hence $f''(x)$ has a root in the interval $(-1,0)$ and a root in the interval $(0,1)$.
Since $x=\pm 1$ are roots of multiplicity $n$ for $f(x)$, $f^{(r)}(\pm 1)=0$ for every $r<n$.