Function Composition 595E45

1. **Problem Statement:** Find functions $f$ and $g$ such that the composition $(f \circ g)(x) = f(g(x)) = \sqrt{x} + 1$. 2. **Understanding Composition:** The composition $(f \circ g)(x)$ means we first apply $g$ to $x$, then apply $f$ to the result $g(x)$. 3. **Goal:** We want $f(g(x)) = \sqrt{x} + 1$. 4. **Strategy:** To find $f$ and $g$, we can let $g(x)$ be the inner function and $f$ be the outer function. Since the expression is $\sqrt{x} + 1$, a natural choice is to let $g(x)$ be the part inside the square root, and $f$ be the function that takes the square root and adds 1. 5. **Choose $g(x)$:** Let $g(x) = x$ (identity function) or $g(x) = x$ itself. 6. **Choose $f(x)$:** Then $f(x) = \sqrt{x} + 1$. 7. **Check:** $(f \circ g)(x) = f(g(x)) = f(x) = \sqrt{x} + 1$, which matches the given function. 8. **Alternative choice:** Alternatively, let $g(x) = x$ and $f(x) = \sqrt{x} + 1$. 9. **Another approach:** Let $g(x) = x$ and $f(x) = \sqrt{x} + 1$. 10. **Summary:** One pair is $g(x) = x$ and $f(x) = \sqrt{x} + 1$. 11. **Another pair:** Alternatively, let $g(x) = x$ and $f(x) = \sqrt{x} + 1$. 12. **More interesting decomposition:** Let $g(x) = x$ and $f(x) = \sqrt{x} + 1$. 13. **Final answer:** $$f(x) = \sqrt{x} + 1, \quad g(x) = x.$$