f\left(x\right)\ =coth^{-1}\left(e^{\sqrt{x'2\ +1}}\right)

The function you've provided is:

\[ f(x) = \coth^{-1}\left(e^{\sqrt{x^2 + 1}}\right) \]

### Simplifying the Expression

1. **Inverse Hyperbolic Cotangent**: The inverse hyperbolic cotangent function \(\coth^{-1}(y)\) is defined as:
   \[
   \coth^{-1}(y) = \frac{1}{2} \ln\left(\frac{y+1}{y-1}\right) \quad \text{for } |y| > 1
   \]

2. **Substitute the Expression**: In your case, \( y = e^{\sqrt{x^2 + 1}} \).
   \[
   f(x) = \frac{1}{2} \ln\left(\frac{e^{\sqrt{x^2 + 1}} + 1}{e^{\sqrt{x^2 + 1}} - 1}\right)
   \]

3. **Further Simplification**:
   \[
   f(x) = \frac{1}{2} \ln\left(\frac{e^{\sqrt{x^2 + 1}}(1 + e^{-\sqrt{x^2 + 1}})}{e^{\sqrt{x^2 + 1}}(1 - e^{-\sqrt{x^2 + 1}})}\right)
   \]

   Simplifying the fraction inside the logarithm:
   \[
   f(x) = \frac{1}{2} \ln\left(\frac{1 + e^{-\sqrt{x^2 + 1}}}{1 - e^{-\sqrt{x^2 + 1}}}\right)
   \]

Thus, the function simplifies to:
\[
f(x) = \frac{1}{2} \ln\left(\frac{1 + e^{-\sqrt{x^2 + 1}}}{1 - e^{-\sqrt{x^2 + 1}}}\right)
\]

Would you like more details or have any questions about this simplification?

Here are 5 related questions:
1. What are the properties of the inverse hyperbolic cotangent function?
2. How can we find the derivative of \( f(x) \) with respect to \( x \)?
3. What is the domain of the function \( f(x) \) provided?
4. How does the function \( f(x) \) behave as \( x \) approaches infinity?
5. Can \( f(x) \) be approximated using a Taylor series expansion?

**Tip**: When working with inverse hyperbolic functions, it is often useful to express them in terms of logarithms to simplify calculations and understand their behavior better.

Understand and simplify the inverse hyperbolic cotangent function \( f(x) = \coth^{-1}\left(e^{\sqrt{x^2 + 1}}\right) \). Learn about the properties, derivative, domain, behavior at infinity, and Taylor series approximation. Suitable for advanced mathematics enthusiasts.

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