A high school problem on derivatives. I came across this problem in an old high school textbook of 1978:

Suppose that $f, g$ are polynomials with domain and image all of $\mathbb{R}$. Prove that if
\begin{equation}
    \begin{cases} 
                  f(x)\neq g(x) \\
                  f''(x)\neq g''(x) 
    \end{cases} \forall \hspace{.1cm} x\in \mathbb{R}
\end{equation}
there is exactly one solution of $f'(x)=g'(x)$
 Clarification: By $f'$ and $f''$, the first and second derivatives are inferred.

I'm not sure what I must do, but I have come up with  a few vague ideas: 

*

*The polynomials can only be of odd order $\geq 3$ and they must have the same coefficient in their largest power. (Otherwise, a new polynomial of odd order would arise, which would oppose the assumption that $f(x)\neq g(x)$)

*I have started by assuming that there is no $x_0$ such that $f'(x_0)=g'(x_0),$ which in turn implies that the function $h(x)=f(x)-g(x)$ is strictly monotone. Can I consider cases? Would that be helpful?

*Also because I need $f''(x)\neq g''(x)$, the function $q(x)=f'(x)-g'(x)$ is also strictly monotone.

There are also other ideas that are just floating around, but I just can't poke the problem well enough. Do you have any ideas?
 A: I don't claim that this is the most elegant solution, but it builds on your own reasoning (I think).
There is at least one solution to $f'(x) = g'(x)$.
The intermediate value theorem implies that either
$$ f(x) - g(x) > 0 \qquad\text{or}\qquad f(x) - g(x) < 0 $$
for all $x$.  In either case, $f-g$ is a real polynomial with no roots, which means that $f-g$ is of even degree (as every real polynomial function of odd degree has at least one real root—this is also a consequence of the intermediate value theorem).  The derivative of a polynomial function of even degree is either

*

*the zero polynomial, or

*a polynomial of odd degree.

But if $(f-g)'$ is the zero polynomial, then $f''(x) = g''(x) = 0$ for all $x$.  Therefore $(f-g)'$ is a polynomial of odd degree.  As noted above, every real polynomial of odd degree has a real root, hence $(f-g)'$ has at least one real root.
(A) Therefore there is at least one real number $x$ such that $ f'(x) = g'(x)$.
There is at most one solution to $f'(x) = g'(x)$.
The intermediate value theorem further implies that either
$$f''(x) - g''(x) > 0 \qquad\text{or}\qquad f''(x) - g''(x) < 0 $$
for all $x$.  This implies that $(f-g)'$ is either strictly increasing, or strictly decreasing (this is a consequence of the mean value theorem).  A strictly monotonic function can have at most one root.
(B) Therefore there is at most one real number $x$ such that $f'(x) = g'(x)$.
From which we conclude...
The two statements (A) and (B) are the desired result.
