'Is $ℜ$ an equivalence relation?

[zeraoulia]

Homework Statement

The **mean value theorem** says that there exists a $c∈(u,v)$ such that $$f(v)-f(u)=f′(c)(v-u).$$ Here is my question.

Assume that $u$ is a root of $f$, hence we obtain $$f(v)=f′(c)(v-u);$$ assume that $f$ is a non-zero analytic function in the whole real line.

We can define the relation $ℜ$ by
$$cℜd⇔f′(c)=\frac{1-t}{1-u}f′(d)$$

where $u∈ℝ$ such that $f(u)=0$, $c∈(u,v)$ and $t∈ℝ$ such that $f(t)=0$, $d∈(t,v)$, that is we apply the mean value theorem for f in the two intervals $(u,v)$ and $(t,v)$, here $u,t,v$ are arbitrary.

Is $ℜ$ an equivalence relation? If so, determine its equivalence classes.

Homework Equations

The Attempt at a Solution

I have no idea to start with.

 

[HallsofIvy]

I am confused by your "Assume that u is a root of f", "assume that f is a non-zero function in the whole real line" and "where u such that f(u)= 0".

 

[mfb]

"non-zero function" is equivalent to "NOT (f(x)=0 everywhere)". And the function is defined in the whole real line.

I am confused by that t, however. Do we have some constant t? Are we free to choose t for each pair (c,d) to analyze? What about d, can we choose that as well?