Consider the function y = 2 / ln((e^cos(x)) + 5). Find functions f, g, h, m and t
such that
y = f ◦ g ◦ h ◦ m ◦ t = f(g(h(m(t(x))))).
Determine the domain of the overall function and state your answer in interval notation

To express \( y = 2 / \ln(e^{\cos(x)} + 5) \) as \( y = f(g(h(m(t(x))))) \), we need to break down the function into a composition of simpler functions. Here's how we can do this:

1. **Innermost function \( t(x) \):** Start with \( t(x) = \cos(x) \).

2. **Next function \( m(t) \):** Apply the exponential function to the result of \( t(x) \):
   \[
   m(t) = e^t
   \]
   Therefore, \( m(t(x)) = e^{\cos(x)} \).

3. **Next function \( h(m) \):** Add 5 to the result of \( m(t) \):
   \[
   h(m) = m + 5
   \]
   So, \( h(m(t(x))) = e^{\cos(x)} + 5 \).

4. **Next function \( g(h) \):** Apply the natural logarithm to the result of \( h(m) \):
   \[
   g(h) = \ln(h)
   \]
   Hence, \( g(h(m(t(x)))) = \ln(e^{\cos(x)} + 5) \).

5. **Outermost function \( f(g) \):** Take the reciprocal of \( g(h) \) and multiply by 2:
   \[
   f(g) = \frac{2}{g}
   \]
   Therefore, \( f(g(h(m(t(x))))) = \frac{2}{\ln(e^{\cos(x)} + 5)} \).

So the functions are:
\[
\begin{aligned}
t(x) &= \cos(x), \\
m(t) &= e^t, \\
h(m) &= m + 5, \\
g(h) &= \ln(h), \\
f(g) &= \frac{2}{g}.
\end{aligned}
\]

Now, the overall function \( y \) can be expressed as:
\[
y = f(g(h(m(t(x))))) = \frac{2}{\ln(e^{\cos(x)} + 5)}.
\]

### Determining the Domain

To find the domain, consider where each function in the composition is defined:

1. **\( t(x) = \cos(x) \):** Defined for all \( x \in \mathbb{R} \).

2. **\( m(t) = e^t \):** Exponential functions are defined for all real numbers, so \( e^{\cos(x)} \) is also defined for all \( x \in \mathbb{R} \).

3. **\( h(m) = m + 5 = e^{\cos(x)} + 5 \):** Addition of 5 to \( e^{\cos(x)} \) is defined for all \( x \in \mathbb{R} \).

4. **\( g(h) = \ln(h) \):** The natural logarithm function is defined for positive values. Therefore, \( e^{\cos(x)} + 5 \) must be greater than 0. Since \( e^{\cos(x)} \) is always positive (it ranges from \( e^{-1} \) to \( e^1 \)), \( e^{\cos(x)} + 5 \) is always positive. Hence, \( \ln(e^{\cos(x)} + 5) \) is defined for all \( x \in \mathbb{R} \).

5. **\( f(g) = \frac{2}{g} \):** We need to ensure that \( g \neq 0 \). Therefore, \( \ln(e^{\cos(x)} + 5) \) must not be zero.

   \[
   \ln(e^{\cos(x)} + 5) = 0
   \]
   implies
   \[
   e^{\cos(x)} + 5 = 1.
   \]

   Since \( e^{\cos(x)} \geq e^{-1} \approx 0.3679 \) and \( e^{\cos(x)} + 5 \geq 5.3679 \), the expression \( e^{\cos(x)} + 5 \) can never be equal to 1. Thus, \( \ln(e^{\cos(x)} + 5) \) is never zero.

Since there are no restrictions making the function undefined, the domain of the overall function is all real numbers:

\[
\text{Domain} = (-\infty, \infty).
\]

Would you like further details or have any questions about this process?

### Related Questions

1. How would the function change if the base of the logarithm were different?
2. What would the derivative of \( y = 2 / \ln(e^{\cos(x)} + 5) \) be?
3. What is the range of the function \( y = 2 / \ln(e^{\cos(x)} + 5) \)?
4. How would the composition change if an additional transformation were applied to \( \cos(x) \)?
5. Can this composition be used to simplify integrals involving this function?

### Tip

When determining the domain of a function, always look for operations that impose restrictions, such as square roots (must be non-negative inside), logarithms (must be positive inside), and denominators (must not be zero).

Consider the function y = 2 / ln((e^cos(x)) + 5). Find functions f, g, h, m and t such that y = f ◦ g ◦ h ◦ m ◦ t = f(g(h(m(t(x))))). Determine the domain of the overall function and state your answer in interval notation.

Explore the composition of functions in y = 2 / ln(e^cos(x) + 5), breaking it down into five simpler functions. Learn how to determine the domain of this complex function and express it in interval notation. Suitable for Calculus I/II students.

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