Number of roots?

Consider the function $h(x) = f'(x) \times f(x)$ . Observe that $h'(x) = g(x)$ . We will study $h(x)$ to understand the 0's of its' derivative $g(x)$ .

By the intermediate value theorem , $f(x) = 0$ for at least one value in $(b,c)$ and in $(c,d)$ . Since $f(a) = 0 , f(e) = 0$ thus $f(x) = 0$ for at least 4 values in $[a,e]$ .

By rolle's theorem , $f'(x) = 0$ for at least 3 values in $[a,e]$ .

Hence, combining these two, we get that $h(x) = f'(x) \times f(x) = = 0$ for at least 7 values (possibly repeated) in $[a,e]$ .

Once again, by rolle's theorem , $h'(x) = 0$ for at least 6 values in $[a,e]$ .

This establishes a lower bound.

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Now, in order to ensure that we have a minimum, let's show that this can be achieved. Looking at the solution, we need that $f(x)=0$ has at most 4 solutions and $f'(x) =0$ has at most 3 solutions. Using interpolation on the point $(0,0), (1,2), (2, -1), (3, 2), (4,0)$ , we obtain the polynomial

$f(x) = - \frac{11}{12} x^4 + \frac{22}{3} x^3 - \frac{217}{12} x^2 + \frac{ 41}{3} x$

It is now clear that $f(x)$ with degree 4 has at most 4 solutions, and $f'(x)$ with degree 3 has at most 3 solutions. Thus we are done.