In Exercises 35-42, find (a) \( f \circ g \) and (b) \( g \circ f \). Find the domain of each function and each composite function. \( \begin{array}{ll}\text { 35. } f(x)=\sqrt{x+4}, & g(x)=x^{2} \\ \text { 36. } f(x)=\sqrt[3]{x-5}, & g(x)=x^{3}+1 \\ \text { 37. } f(x)=x^{2}+1, & g(x)=\sqrt{x} \\ \text { 38. } f(x)=x^{2 / 3}, & g(x)=x^{6}\end{array} \)

Let's solve each exercise step by step. ### Exercise 35 **Functions:** - \( f(x) = \sqrt{x + 4} \) - \( g(x) = x^2 \) **(a) Find \( f \circ g \):** \[ f \circ g(x) = f(g(x)) = f(x^2) = \sqrt{x^2 + 4} \] **Domain of \( f \circ g \):** - The function \( g(x) = x^2 \) is defined for all \( x \). - The output of \( g(x) \) is \( x^2 \), which is always non-negative. Thus, \( x^2 + 4 \geq 4 \) is always valid. - Therefore, the domain of \( f \circ g \) is all real numbers: \( (-\infty, \infty) \). **(b) Find \( g \circ f \):** \[ g \circ f(x) = g(f(x)) = g(\sqrt{x + 4}) = (\sqrt{x + 4})^2 = x + 4 \] **Domain of \( g \circ f \):** - The function \( f(x) = \sqrt{x + 4} \) is defined when \( x + 4 \geq 0 \) or \( x \geq -4 \). - Therefore, the domain of \( g \circ f \) is \( [-4, \infty) \). --- ### Exercise 36 **Functions:** - \( f(x) = \sqrt[3]{x - 5} \) - \( g(x) = x^3 + 1 \) **(a) Find \( f \circ g \):** \[ f \circ g(x) = f(g(x)) = f(x^3 + 1) = \sqrt[3]{(x^3 + 1) - 5} = \sqrt[3]{x^3 - 4} \] **Domain of \( f \circ g \):** - The function \( g(x) = x^3 + 1 \) is defined for all \( x \). - The output \( x^3 - 4 \) can take any real value, so \( f \circ g \) is defined for all \( x \). - Therefore, the domain of \( f \circ g \) is \( (-\infty, \infty) \). **(b) Find \( g \circ f \):** \[ g \circ f(x) = g(f(x)) = g(\sqrt[3]{x - 5}) = (\sqrt[3]{x - 5})^3 + 1 = x - 5 + 1 = x - 4 \] **Domain of \( g \circ f \):** - The function \( f(x) = \sqrt[3]{x - 5} \) is defined for all \( x \). - Therefore, the domain of \( g \circ f \) is \( (-\infty, \infty) \). --- ### Exercise 37 **Functions:** - \( f(x) = x^2 + 1 \) - \( g(x) = \sqrt{x} \) **(a) Find \( f \circ g \):** \[ f \circ g(x) = f(g(x)) = f(\sqrt{x}) = (\sqrt{x})^2 + 1 = x + 1 \] **Domain of \( f \circ g \):** - The function \( g(x) = \sqrt{x} \) is defined for \( x \geq 0 \). - Therefore, the domain of \( f \circ g \) is \( [0, \infty) \). **(b) Find \( g \circ f \):** \[ g \circ f(x) = g(f(x)) = g(x^2 + 1) = \sqrt{x^2 + 1} \] **Domain of \( g \circ f \):** - The function \( f(x) = x^2 + 1 \) is defined for all \( x \). - The output \( x^2 + 1 \) is always non-negative, so \( g \circ f \) is defined for all \( x \). - Therefore, the domain of \( g \circ f \) is \( (-\infty, \infty) \). --- ### Exercise 38 **Functions:** - \( f(x) = x^{2/3} \) - \( g(x) = x^6 \) **(a) Find \( f \circ g \):** \[