Suppose \( H(x)=5 \sqrt{x}-3 \), 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.) \( f(x)=\square \) \( g(x)=\square \)

One way to express \( H(x) = 5 \sqrt{x} - 3 \) as the composition of two functions is to break it down into two simpler parts. Let's choose: \( g(x) = x \) (which is not allowed, but let's start here to simplify our thinking), then, for \( f(x) \), we can define it as: \( f(x) = 5\sqrt{x} - 3 \). However, since both \( f \) and \( g \) must not be identity functions, we revise our \( g \): Let \( g(x) = x^2 \) (this is a valid non-identity function), then: \( f(x) = 5\sqrt{x} - 3 \) becomes \( f(g(x)) = 5\sqrt{x^2} - 3 = 5x - 3 \). To fit better, we can choose: \( g(x) = \frac{x + 3}{5} \), then: \( f(x) = 5\sqrt{x} - 3 \) leads back to H since \( f(g(x)) = 5\sqrt{\frac{x + 3}{5}} - 3 \). Thus, one valid solution could be: \( f(x) = 5 \sqrt{x} - 3 \) and \( g(x) = x^2 \). But let's take a concrete non-identity formulation for both: \( f(x) = 5x - 3 \), \( g(x) = x^{2/5} \). So to summarize, here are the functions: \( f(x) = 5x - 3 \) \( g(x) = x^{2/5} \)