Solution:
We are given two Hyperbolic Functions \( \sinh x \) and \( \cosh x \)
1. Understanding the Hyperbolic Functions:
The definitions of the hyperbolic functions are:
- \( \sinh x = \frac{e^x - e^{-x}}{2} \)
- \( \cosh x = \frac{e^x + e^{-x}}{2} \)
2. Finding the Domain:
Both \( \sinh x \) and \( \cosh x \) are defined for any real value of \( x \) because the exponential function is always defined.
Thus, the domain of both functions is:
\[ \text{Domain} = (-\infty, \infty) \]
3. Finding the Range of \( \sinh x \):
Since the exponential function \( e^x \) always returns positive values, the expression \( \sinh x = \frac{e^x - e^{-x}}{2} \) will cover all real values as \( x \) varies.
- As \( x \) approaches positive infinity: \[ \sinh x \approx \frac{e^x}{2} \rightarrow \infty \]
- As \( x \) approaches negative infinity: \[ \sinh x \approx -\frac{e^{-x}}{2} \rightarrow -\infty \]
4. Finding the Range of \( \cosh x \):
The hyperbolic cosine function \( \cosh x = \frac{e^x + e^{-x}}{2} \) is always positive because it sums two positive terms.
It's symmetric about the y-axis (even function) because:
\[ \cosh(-x) = \cosh x \]
The minimum value of \( \cosh x \) is 1 when \( x = 0 \):
\[ \cosh(0) = \frac{e^0 + e^{-0}}{2} = 1 \]
As \( x \) approaches positive or negative infinity, \( \cosh x \) increases exponentially:
\[ \cosh x \approx \frac{e^x}{2} \rightarrow \infty \]
Therefore, the range of \( \cosh x \) is:
\[ \text{Range of } \cosh x = [1, \infty) \]