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Let we have a function:

$f(x)=(x-1)(x-2)(x-3)(x-4)(x-5)$

The question is: There are how many distinct real roots of the polynomial $\frac{d}{dx}f(x)=0$ has?.

My approach:

Clearly, $f(x)=0$ has five roots. So, we can say that the derivative of this function will have at least four roots. The answer is four. On plotting this function on graph, it turns out that this function is parallel to x-axis four times. But, my question is that can we prove that without using graph. One way is to calculate the derivative, but it will be very tedious to calculate the derivative of this function. Using the rolle's theorem, we can only that the $f'(x)$ will have at least four roots. We cannot be precise (Correct me if I am wrong). My question is that can we prove that without using graph

Thanks.

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First the number of distinct real roots must be less than or equal to $4$ by the degree of the polynomial.

Then for any two points (say $1$ and $2$, you can easily generalize this to other points), we can apply Rolle's Theorem to show that, since $f(2)-f(1)=0$, so there exists a point $x$ with $f'(x)=0$, and $1<x<2$. So we are done (there are $4$ distinct points with $f'(x)=0$ as required).

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First, the derivative cannot have more than 4 roots.

Next, the function itself changes its sign five times (consider its sign in between and beyond its roots).

In particular, there are four distinct (adjoint) segments such that, on each segment, the function is non-zero in any inner point and zeroes on both ends; since it is a smooth curve, it has to have at least one extremum inside each one. Since there are four such distinct segments, there are four roots to the derivative.

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