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:

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).