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 handle the inner part of the expression, and \( f(x) \) to handle the outer part. Let's define: \[ g(x) = 4(x^2 + 1) \] This function takes \( x \) and transforms it into \( 4(x^2 + 1) \). Next, we need to define \( f(x) \) to take the output of \( g(x) \) and apply the cube root: \[ f(x) = \sqrt[3]{x} \] Now, we can check the composition: \[ (f \circ g)(x) = f(g(x)) = f(4(x^2 + 1)) = \sqrt[3]{4(x^2 + 1)} = \sqrt[3]{4x^2 + 4} \] Thus, we have: \[ (f \circ g)(x) = H(x) \] Therefore, one possible pair of functions is: \[ f(x) = \sqrt[3]{x}, \quad g(x) = 4(x^2 + 1) \] Another possible pair could be to define \( g(x) \) differently. For example, we could define: \[ g(x) = x^2 + 1 \] and then adjust \( f(x) \) accordingly: \[ f(x) = \sqrt[3]{4x} \] Now, checking this composition: \[ (f \circ g)(x) = f(g(x)) = f(x^2 + 1) = \sqrt[3]{4(x^2 + 1)} = \sqrt[3]{4x^2 + 4} \] Thus, another valid pair of functions is: \[ f(x) = \sqrt[3]{4x}, \quad g(x) = x^2 + 1 \] In summary, two pairs of functions that satisfy the requirement are: 1. \( f(x) = \sqrt[3]{x}, \quad g(x) = 4(x^2 + 1) \) 2. \( f(x) = \sqrt[3]{4x}, \quad g(x) = x^2 + 1 \)