Suppose \( H(x)=\sqrt{4 x^{2}+4} \) 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 \( (f \circ g)(x) = H(x) = \sqrt{4x^2 + 4} \), we can start by simplifying \( H(x) \). First, we can factor out the constant inside the square root: \[ H(x) = \sqrt{4(x^2 + 1)} = \sqrt{4} \cdot \sqrt{x^2 + 1} = 2\sqrt{x^2 + 1} \] Now, we can express \( H(x) \) as a composition of two functions. One possible way to do this is to let \( g(x) \) be a function that transforms \( x \) into \( x^2 + 1 \), and then let \( f(y) \) be a function that takes \( y \) and outputs \( 2\sqrt{y} \). Let's define the functions as follows: 1. Let \( g(x) = x^2 + 1 \). 2. Let \( f(y) = 2\sqrt{y} \). Now we can check the composition: \[ (f \circ g)(x) = f(g(x)) = f(x^2 + 1) = 2\sqrt{x^2 + 1} \] This gives us: \[ (f \circ g)(x) = 2\sqrt{x^2 + 1} = H(x) \] Thus, we have found one pair of functions \( f \) and \( g \): \[ f(y) = 2\sqrt{y}, \quad g(x) = x^2 + 1 \] Another possible pair could be: 1. Let \( g(x) = 2x^2 + 2 \). 2. Let \( f(y) = \sqrt{y} \). Now we can check this composition: \[ (f \circ g)(x) = f(g(x)) = f(2x^2 + 2) = \sqrt{2x^2 + 2} = \sqrt{2(x^2 + 1)} = \sqrt{2} \cdot \sqrt{x^2 + 1} \] This does not yield \( H(x) \) directly, but we can adjust \( g(x) \) to ensure it fits. Instead, we can also try: 1. Let \( g(x) = 4x^2 + 4 \). 2. Let \( f(y) = \sqrt{y} \). Now we check: \[ (f \circ g)(x) = f(g(x)) = f(4x^2 + 4) = \sqrt{4x^2 + 4} = H(x) \] Thus, we have another valid pair: \[ f(y) = \sqrt{y}, \quad g(x) = 4x^2 + 4 \] In conclusion, two pairs of functions \( f \) and \( g \) that satisfy \( (f \circ g)(x) = H(x) \) are: 1. \( f(y) = 2\sqrt{y}, \quad g(x) = x^2 + 1 \) 2. \( f(y) = \sqrt{y}, \quad g(x) = 4x^2 + 4 \)