A Problem with Intermediate Value Property and a partial converse question An exercise in real analysis asks to show if $f:[0,1] \to \mathbb{R}$ is a function such that $f(0)<0<f(1)$ and there is a continuous function $g:[0,1]\to \mathbb{R}$ such that $f+g$ is decreasing, then there is an $\omega \in (0,1)$ such that $f(\omega)=0$. 
Now my question is how to link $\exists g\in \mathrm{C}[0,1]$, such that $f+g$ is strictly monotone $\implies$ $f$ has I.V.P. ?
My second question is if it is true that for any function $f$ satisfying the Intermediate value property in $[0,1]$, there is a continuous function $g$ in $[0,1]$ such that $f+g$ is strictly monotone in a sub interval of $[0,1]$ ?
I ask the second question to know if there is a converse to the first question.
 A: The following fact will be useful:

Fact. Let $I$ be an interval, and let $h: I \to \mathbb{R}$ be a monotone function.  Then for any $a$, $\lim_{x \to a^+}$ and $\lim_{x \to a^-}$ exist.

Question 1

Show if $f:[0,1] \to \mathbb{R}$ is a function such that $f(0)<0<f(1)$ and there is a continuous function $g:[0,1]\to \mathbb{R}$ such that $f+g$ is decreasing, then there is an $\omega \in (0,1)$ such that $f(\omega)=0$. 

Let $h(x) = f(x) + g(x)$
be decreasing.
Let
$$
a = \sup \{ x \; : \; f(x) < 0\}
$$
Since $h$ is decreasing, we have
$$
\lim_{x \to a^-} h(x) \ge h(a) \ge \lim_{x \to a^+} h(x)
$$
Since $g$ is continuous and $f = h - g$, the above implies $\lim_{x \to a^+} f(x)$ and $\lim_{x \to a^-} f(x)$ exist, and
$$
\lim_{x \to a^-} f(x) \ge f(a) \ge \lim_{x \to a^+} f(x)
$$
Now, by definition of $a$,
we must have $\lim\limits_{x \to a^+} f(x) \ge 0$.
This implies $f(a) \ge 0$,
so again by definition of $a$,
there must be a sequence $x_n$ increasing to $a$ such that
$f(x_n) < 0$, so
$\lim\limits_{x \to a^-} f(x) \le 0$.
Therefore,
$$
0 \ge \lim_{x \to a^-} f(x) \ge f(a) \ge \lim_{x \to a^+} f(x) \ge 0.
$$
Question 2

If it is true that for any function $f$ satisfying the Intermediate value property in $[0,1]$, there is a continuous function $g$ in $[0,1]$ such that $f+g$ is strictly monotone in a sub interval of $[0,1]$ ?

Assuming I am interpreting your question correctly, the answer is no.
By the fact at the start of this answer,
a monotone function and a continuous function add to get a function such that
the left and right limit exist everywhere, i.e.
the function only has jump disconinuities.
But on the other hand,
some functions which satisfy the intermediate value property do not have a
left and right limit everywhere.
For example, take the famous example of 
$$
f(x) =
\begin{cases}
\sin(1/x) & \text{if } x > 0 \\
0 & \text{if } x = 0.
\end{cases}
$$
Functions that satisfy the intermediate value property are called "Darboux functions",
and they can be extremely irregular.
In particular, a Darboux function has no jump discontinuities, but only essential discontinuities (one of the left and right limits must not exist).
This fact alone says that the converse (i.e. Question 2) must be false, for if not then all Darboux functions would be continuous.
