Solution: We will prove that such a function does not exist by showing the following.
-
$h$ is surjective.
-
$f$ is strictly monotone function.
-
$f$ is bijective.
Since $g$ is continuous, without loss of generality, we can assume that $g(0)= 0$. Note that if the above are true, then for any $x \in \mathbb{R} $, there exists $y \in \mathbb{R} $ such that $f(y) = x$. Thus,
\[
h(x) = h(f(y) + g(0)) = 0,
\]
which implies $h\equiv 0$, a contradiction. Thus, there does not exist such continuous functions.
Now let us show the above claims.
-
It is clear that $h$ is surjective, by taking $x \in \mathbb{R} $ and $y = 1$.
-
Suppose that $f(x) = f (y)$ for $x, y \in \mathbb{R} $. Then for any $z \in \mathbb{R}\setminus \{ 0 \} $
\begin{align*}
& h(f(x) + g(z)) = h \left( f(y) + g(z) \right) \\
\implies & xz = yz \implies x = y.
\end{align*}
Thus, $f$ is injective function and hence $f$ is strictly monotone function.
-
Now note that if $f$ is bounded above, then $\displaystyle \lim_{x \to \infty} f(x) = \alpha $ for some real number $\alpha $. Thus, using the continuity of $h$ we have
\begin{align*}
h(\alpha + g(1)) & = \lim_{x \to \infty} h\left( f(x) + g(1) \right) \lim_{x \to \infty} x = \infty ,
\end{align*}
a contradiction. Similarly, if $f$ is bounded below, then $\lim_{x \to -\infty } f(x) = \beta $. This implies,
\begin{align*}
h(\beta + g(1)) & = \lim_{x \to -\infty} h\left( f(x) + g(1) \right) \lim_{x \to -\infty} x = -\infty ,
\end{align*}
a contradiction. Thus, $f$ is unbounded injective function. Hence, $f$ is bijective function.