Showing that implicit function is continuous For $x \in (1, \infty)$, define $f(x)$ to be the unique positive solution to the equation $a = \tanh(xa)$.
I am trying to understand basic properties of $f$, in particular, if $f$ is continuous or differrntable.
I think that $f$ is continuous,and the reason is the following: let $x_0 \in (1, \infty)$ be the solution to $a = \tanh(a x_0)$. Then for small $\epsilon$, $\tanh(a(x_0 + \epsilon)) = \frac{\tanh(ax_0) + \tanh(a\epsilon)}{1 + \tanh(ax_0)\tanh(a\epsilon)}$, and as $\epsilon \rightarrow 0$, we get that both the RHS and the LHS are equal by the definition of $x_0$.
However I am not sure how to make this argument formal, or even if this argument is correct.
 A: I would proceed as follows: First fix $x > 1$ and consider the function
$$
g: [0, \infty) \to \Bbb R, \, g(a) = \tanh(xa) - a \, .
$$
The following properties are straightforward to verify:

*

*$g(0) = 0$, $g(1) < 0$,

*$g'(0) > 0$,

*$g$ is strictly concave.

It follows that $g$ has a maximum at some point $a_0 \in (0, 1)$ with $g(a_0) > 0$, $g$ is strictly increasing on $[0, a_0]$ and strictly decreasing on $[a_0, \infty)$.
Consequently, the equation $g(a) = 0$ has exactly one positive solution $a^*$ and we can define $f(x) = a^*$. Note that $a^* \in (0, 1)$ and $g'(a^*) < 0$.
Now we can apply the implicit function theorem to
$$
 F: (1, \infty) \times (0, 1) \to \Bbb R, \, F(x, a) = \tanh(xa) - a\, .
$$
We already know that
$$
 F(x, f(x)) = 0
$$
and
$$
 \frac{\partial}{\partial a} F(x, f(x)) < 0
$$
for all $x > 1$. The implicit function theorem then states that $f$ is differentiable, with
$$
 f'(x) = -\frac{\frac{\partial}{\partial a} F(x, f(x))}{\frac{\partial}{\partial x} F(x, f(x))} \, .
$$
