Let $f(x) = (x-a)(x-b)^3(x-c)^5(x-d)^7$, where $a,b,c,d \in \mathbb{R}$ and $a<b<c<d$. Thus $f(x)$ has 16 real roots counting multiplicities and among them 4 are distinct from each other. Find: (i) the number of distinct real roots of $f'(x)$ (ii) The number of real roots of $f(x)$, counting multiplicities.

Here's what I've tried:

Since $f(a) = f(b) = f(c) = f(d) = 0$, applying Rolle's Theorem on $[a,b], [b,c], [c,d]$ shows that $f(x)$ has 1 root each in the 3 closed intervals, showing that $f'(x)$ has 3 distinct roots.

For the second part, differentiating $f(x)$ gives $$f'(x) =(x-b)^2(x-c)^4(x-d)^6((x-b)(x-c)(x-d)+3(x-a)(x-c)(x-d)+5(x-a)(x-b)(x-d)+7(x-a)(x-b)(x-c))$$ from where we can see that $f'(x)$ has $12$ real roots, including mutiplicites. (We can also observe that $f'(x)$ has 3 distinct roots).

However, the answer to this question is (i) 6, (ii) 15. Is there anything I am doing wrong?

  • $\begingroup$ You can see that f'(x) has 3 distinct roots b,c,d and three other distinct roots on each open interval (a,b);(b,c);(c,d) $\endgroup$ – HLong Sep 28 '14 at 5:37

There are 12 "obvious" roots, three of them distinct, at $x=b,c,d$. There must also be one root (as you say, by Rolle's theorem) in each of the three intervals $(a,b)$, $(b,c)$, and $(c,d)$. This makes for 15 total roots (6 of them distinct) and since $f'$ has degree 15 there can't be any others.

  • $\begingroup$ ah, I totally forgot that the root is between $(a,b)$, and not $[a,b]$ :| $\endgroup$ – Vibhav Pant Sep 28 '14 at 5:46

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