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Question 10: Find the Domain and Range of a Function,$$ f(x) =\sinh x $$ and $$ f(x) =\cosh x $$

Solution:

We are given two Hyperbolic Functions \( \sinh x \) and \( \cosh x \)

1. Understanding the Hyperbolic Functions:
The definitions of the hyperbolic functions are:

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.

Therefore, the range of \( \sinh x \) is: \[ \text{Range of } \sinh x = (-\infty, \infty) \]

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: \[ \text{Range of } \cosh x = [1, \infty) \]