<s>[INST] Number of zeros of $f(x)= \frac{1}{2} E\left[ \tanh \left( \frac{x+Z}{2} \right) \right]-\tanh(x)+\frac{x}{2}$ where $Z$ is standard normal Consider the following function:
\begin{align}
f(x)= \frac{1}{2} E\left[ \tanh \left( \frac{x+Z}{2} \right) \right]-\tanh(x)+\frac{x}{2},
\end{align}
where $Z$ is standard normal.
Question: How to show that this function has only three zeros? Note, that we are not interested in the locations just the number of zeros.
By using that $\tanh(x)$ is an odd function, it is not difficult to show that $f(0)=0$.
However, I am not sure how to show the existence of the other two zeros.
I know that two more zeros exist from the numerical simulation (see the attached figure). 
Edit:  The current answer shows that there are at least 3 zeros.  Now we need to show that there can be no more than 3 zeros.
Edit 2 Idea for a proof.  Consider only positive $x$.  I think  if we can show the following:

*

*$f(x)>0$ for  all $x>x_1$,

*$f(x)$ is convex for $x \in (0,x_2)$, and

*$x_2>x_1$.

Then this will imply that the function is convex in the regime while it changes a sign. Therefore, it can only have at most one sign change.
 [/INST] Not a complete proof, but rather a sketch. First, observe that as $x\to +\infty$, than $\tanh (x) \to 1$, so $f(x)$ asymptotically asymptotically equivalent to $\frac{1}{2} E [1] - 1 + \frac{x}{2} = \frac{x-1}{2}$, so for some large enough positive value of $x$ one has $f(x) > 0$. After that, let's look at the graph of the function and notice that $f'(0)$ must be negative, which is not hard to prove: As
$$
\frac{1}{2} E\left[ \tanh \left( \frac{x+Z}{2} \right) \right] = 
\frac{1}{2} \int_\mathbb{R} \tanh \left( \frac{x+z}{2} \right) \frac{1}{\sqrt{2\pi}} e^{-z^2/2}dz
$$
So
$$
f'(x) = \frac{1}{2} \int_\mathbb{R} 
\frac{\partial}{\partial x}\left[
\tanh \left( \frac{x+z}{2} \right)\right] \frac{1}{\sqrt{2\pi}} e^{-z^2/2}dz + 
\frac{d}{dx} \left[-\tanh (x) + \frac{x}{2}\right] = \\ =
\frac{1}{4} \int_\mathbb{R} \frac{1}{\cosh^2 \left( \frac{x+z}{2} \right)} \frac{1}{\sqrt{2\pi}} e^{-z^2/2}dz - \frac{1}{\cosh^2 (x)} + \frac{1}{2} \\
f'(0) = \frac{1}{4} \int_\mathbb{R} \frac{1}{\cosh^2 \left( \frac{z}{2} \right)} \frac{1}{\sqrt{2\pi}} e^{-z^2/2}dz - 1 + \frac{1}{2} \leq
\frac{1}{4} \int_\mathbb{R} \frac{1}{\sqrt{2\pi}} e^{-z^2/2}dz - \frac{1}{2} = 
\frac{1}{4} - \frac{1}{2} = -\frac{1}{4}
$$
where the inequality is obtained from $\frac{1}{\cosh^2 \left( \frac{z}{2} \right)} \leq 1 \;\forall z \in \mathbb{R}$. So, it means that $f(x)$ is decreasing in some neighborhood of zero.
So, there exists some $x^+ > 0$ s.t. $f(x^*) > 0$ and some $x^- > 0$ s.t. $f(x^-) < f(0) = 0$. So, by intermediate value theorem (it's also an exercise to check the continuity of $f(x)$) there exists some $x^* \in [x^-, x^+]$ s.t. $f(x^*) = 0$. As $f(x)$ is an odd function, it means that $-x^*$ is also a root.
EDIT: as $\frac{1}{2} - \frac{1}{\cosh^2(x)} \leq f'(x) \leq \frac{3}{4} - \frac{1}{\cosh^2(x)}$ and $f'(0) = 0$, I suppose that with similar methods one can prove that $f$ also has a stationary point somewhere on positive axis, which can help in proving that there are no other roots.
</s>