Find the Image of $\tanh{x}$

Question

Problem

Find the Image of $\tanh{x}$

Attempt

I know that $\tanh{x} \rightarrow 1$ for $x \rightarrow \infty$ and $\tanh{x} \rightarrow -1$ for $x \rightarrow -\infty$. I know that $\tanh{x}$ is continuous over the real numbers and I know that I have to argue from the intermediate value thereom. However, I do not know how to apply the IVT to show that the image is indeed $(-1,1)$. I now that I have to pick an element $y\in (-1,1)$ and....

This is where I am stuck. I do not see how the argue for the double inclusion for

$$\tanh{((-\infty,\infty))}=((-1,1))$$

Answer 1

Hint: You can use the fact that $\tanh x$ is an increasing function: $$ (\tanh x)'=\frac1{\cosh^2 x}>0. $$

Answer 2

hint

$$\tanh(x)=\frac{e^x-e^{-x}}{e^x+e^{-x}} $$

$$=1-\frac{2}{e^{2x}+1}=y$$

$$\implies x=\frac 12\ln(\frac{1+y}{1-y})$$

Answer 3

As

$$\tanh x=\frac{e^x-e^{-x}}{e^x+e^{-x}}=\frac{e^{2x}-1}{e^{2x}+1},$$

$$\frac{1+\tanh x}{1-\tanh x}=e^{2x}>0$$ and this occurs for $-1<\tanh x<1$ by resolution of the inequation. And for all $\tanh x$ in that range, $x$ is defined.

Answer 4

Here's the intended argument:

Clearly, the range of $\tanh$ is a subset of $(-1, 1)$, since $|e^x - e^{-x}| < |e^x + e^{-x}|$.

Now, let us show that $(-1, 1)$ is a subset of the range of $\tanh$.

You have already stated that $\lim\limits_{x \to \infty} \tanh(x) = 1$ and $\lim\limits_{x \to -\infty} \tanh(x) = -1$. You also know that $\tanh$ is continuous.

Consider an arbitrary $y \in (-1, 1)$. Then $-1 < y$. Then $y + 1 > 0$. Define $\epsilon = y + 1$. Then consider some $k$ such that for all $x < k$, we have $|\tanh(x) - (-1)| < \epsilon$. Such a $k$ exists by the definition of $\lim\limits_{x \to -\infty}$.

Then in particular, for $x < k$, we have $\tanh(x) - (-1) < \epsilon = y + 1$. Then $\tanh(x) < y$. Take the particular value $x = k - 1$; then $\tanh(x) < y$.

Similarly, $y < 1$. A completely analogous argument allows us to find some $z$ such that $\tanh(z) > y$.

So we have $\tanh(x) < y < \tanh(z)$. Thus, by the intermediate value theorem, we must have some $j$ such that $\tanh(j) = y$.

Therefore, we see that the range of $\tanh$ is indeed $(-1, 1)$.