So then $(f(x)-1)(f(x)+1)f(x)=0$ has 3 roots. Because between $f(x)-1$, $f(x)+1$ and $f(x)$ there are an odd number of roots, either $f(x)-1$, $f(x)+1$ or $f(x)$ has a double root. It must be $f(x)$ though, because if it were $f(x)-1$ then either $f(x)+1$ and $f(x)$ each would have no roots or 2 roots, and similarly if it were $f(x)+1$. So $f(x)$ has a double root, so 0 is the $y$ coordinate of the vertex.

The question is stated incorrectly: for the trinomial $f(x)=x^5+x+1$, $f(x)^3-f(x)=0$ has three real roots and two vertices in $x=-0.919\dots$ and $x=- 0.410\dots$! Moreover, for $f(x)=x^5+x+2$ there are three real roots and other two different vertices. Maybe with "trinomial" do you mean 'a third-degree polynomial' rather than 'a polynomial with three terms'?

For many, a trinomial is just a second degree polynomial $ax^2 + bx + c$ (a quadratic trinomial); probably that was the intended meaning ...

Thank you mavropnevma for the clarification. In this case, the equation becomes $f(x)(f(x)+1)(f(x)-1)=0$, and since it has an odd number of roots (three), at least one of the factors has an odd number of roots as well. If it was $f(x)+1$, then the other two would have two roots (for a positive leading coefficient) or none (if negative), and similarly for $f(x)-1$, so $f(x)$ must have a double zero and therefore a vertex with ordinate $0$.

Hi Cassius, Can you please explain that part "then the other two would have two roots (for a positive leading coefficient) or none (if negative), and similarly for $f(x)-1$, so $f(x)$ must have a double zero and therefore a vertex with ordinate $0$." And also please tell if an equation has double zero, how it means that vertex with ordinate $0$? Please help me understanding these two points.

Sorry to bring this up but can anyone answer the question sengupta asked pls ?

$ (f(x))^3-f(x)=0=f(x)(f(x)+1)(f(x)-1)$. It follows that one of these factors must have 2 roots, while a second must have zero, and the third must have none. Since $-1<0<1$, it must be $f(x)$ that has the double root, hence the y vertex is zero.

By the picture, we can deduce the y coordinate of the vertex of $f(x)$ is $0\,$.

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Nice one! If we factor, $f(x)\cdot(f(x)-1)(f(x)+1)=0$, since, this has $3$, roots. Now, one of these factors needs to have a double root. But this double root, must be $f(x)$, hence, $\boxed{0}$, is the $y$ coordinate.