What is the range of this function: (sin(x)+tanh(x))/(1+sin(x)tanh(x))?

Question

I don't really have any idea how to solve this problem. It appears to have horizontal asymptotes at values of $x=1$ and $x=-1$ but I have no way of proving these. I have taken the derivative (an unwieldy thing) and i don't see any way of solving it for zero without the use of a program. Any solution or insight is welcome and greatly appreciated.

Answer 1

Both $\sin , \tanh $ are bounded within same range $ \pm 1$ and both have common domain $ -\infty < x < \infty $. With such sign combination considering numerator/ denominator it will not affect the common range $\pm 1.$

Answer 2

The denumerator is never zero, so this function is continuous. Also by a simple check you can see that it is also differentiable. By solving $f'(x) = 0$ you will find that the global extrema happen when $\sin x = \pm 1$, where $f(x) = \pm 1$. By continuity the range is $[-1, 1]$.

Answer 3

Since $\tanh(a+b) =\frac{\tanh(a)+\tanh(b)}{1+\tanh(a)\tanh(b)} $, if $\sin(x) =\tanh(y) $, this is $\frac{\tanh(y)+\tanh(x)}{1+\tanh(y)\tanh(x)} =\tanh(x+y) =\tanh(x+\tanh^{(-1)}(\sin(x))) $.

Since the range of $\sin(x)$ is $[-1, 1]$, the range of $\tanh^{(-1)}(\sin(x))$ is $(-\infty, \infty)$ as is the range of $x+\tanh^{(-1)}(\sin(x))$.

Therefore the range of $\tanh(x+\tanh^{(-1)}(\sin(x))) $ is $(-\infty, \infty)$.

If the value for $x = \pi/2$ is considered to be $\infty$, we might allow $\infty$ to be reached.

I found seeing the plots of the various components (in Wolfy) to be interesting.