Suppose \( H(x)=\sqrt[3]{4 x^{2}+4} \). Find two functions \( f \) and \( g \) such that \( (f \circ g)(x)=H(x) \). Neither function can be the identity function. (There may be more than one correct answer.)

To find two functions \( f \) and \( g \) such that \( (f \circ g)(x) = H(x) = \sqrt[3]{4x^2 + 4} \), we can start by breaking down the expression for \( H(x) \). We can express \( H(x) \) in a way that suggests a composition of functions. Notice that we can factor out a 4 from the expression inside the cube root: \[ H(x) = \sqrt[3]{4(x^2 + 1)} \] Now, we can define \( g(x) \) to be the inner function that prepares the input for \( f \). A natural choice for \( g(x) \) is to take the expression inside the cube root: \[ g(x) = 4(x^2 + 1) \] Next, we need to define \( f(y) \) such that it takes the output of \( g(x) \) and produces \( H(x) \). Since \( g(x) \) outputs \( 4(x^2 + 1) \), we can define \( f(y) \) as the cube root function: \[ f(y) = \sqrt[3]{y} \] Now we can check the composition: \[ (f \circ g)(x) = f(g(x)) = f(4(x^2 + 1)) = \sqrt[3]{4(x^2 + 1)} = H(x) \] Thus, we have found two functions: \[ g(x) = 4(x^2 + 1) \quad \text{and} \quad f(y) = \sqrt[3]{y} \] These functions satisfy the requirement that \( (f \circ g)(x) = H(x) \) and neither function is the identity function. In conclusion, one possible pair of functions is: \[ f(y) = \sqrt[3]{y}, \quad g(x) = 4(x^2 + 1) \]