Let $$f(x)=(x-a)(x-b)^3(x-c)^5(x-d)^7 $$
where $a,b,c,d$
are real numbers with
$a < b < c < d$
. Thus $ f
(
x
)$ has $16$ real roots counting multiplicities and among them $4$ are
distinct from each other. Consider
$f
'
(
x
)$, i.e. the derivative of
$f
(
x
)$. Find the following:
$(i)$ the number of real roots of
$f
'
(
x
)$, counting multiplicities,
$(ii)$ the number of
distinct
real roots of
$f
'
(
x
)$.
This is a polynomial of degree $16$ hence the derivative will be of degree $15$ and hence it will have $15$ roots. But are they real ?
How to find distinct real roots ? Rolle's theorem tells only about existence of root.
If $p(x) = (x-\lambda)^n q(x)$ then from the product rule:
$p'(x) = n(x-\lambda)^{n-1}q(x)+(x-\lambda)^nq'(x)=(x-\lambda)^{n-1}\left[nq(x)+(x-\lambda)q'(x)\right]$
This shows that if $\lambda$ is a root of $p(x)$ of multiplicity $n>1$, then $\lambda$ is also a root of $p'(x)$ with multiplicity $n-1$.
This gets you 2 + 4 + 6 = 12 real roots from b, c, and d, and the other 3 come from Rolle's theorem.
To summarize: 15 real roots, 6 of which are distinct.