Question

 For \(x \in \mathbb{R}\), two real‐valued functions \(f(x)\) and \(g(x)\) are such that

Answer 1

We are given \[ g(x) = \sqrt{x} + 1, \quad f(g(x)) = x + 3 - \sqrt{x}. \] Notice that wherever we see \(g(x)\) in an expression for \(f \circ g\), we can treat that as a single variable, say \(u = g(x)\). Then \[ f(u) = u^2 - 3u + 5, \] because substituting \(u = g(x)\) into \(u^2 - 3u + 5\) recovers the given formula: \[ (u)^2 - 3(u) + 5 = \bigl(\sqrt{x}+1\bigr)^2 - 3\bigl(\sqrt{x} + 1\bigr) + 5 = x + 2\sqrt{x} + 1 - 3\sqrt{x} - 3 + 5 = x - \sqrt{x} + 3, \] exactly matching the right‐hand side. Hence, \(f(x) = x^2 - 3x + 5\). Therefore, \[ f(0) = 0^2 - 3\cdot 0 + 5 = 5. \] Note on domain: Strictly speaking, for \(f \circ g\) to be defined at \(x=0\), \(g(0)\) must lie in the domain of \(f\). Here \(g(0) = 1\), which poses no difficulty for the form \(f(x)= x^2 - 3x + 5\). Thus the final result is \(f(0)=5\).