Key Concepts and Formulas

• Composition of Functions: For two functions $f$ and $g$, the composite function $f \circ g(x)$ is defined as $f(g(x))$.
• Algebraic Manipulation: The ability to rearrange equations and substitute expressions is crucial for solving problems involving function composition.
• Domain and Range Considerations: While not explicitly required for this problem's calculation, understanding the domains and ranges of functions is essential for a complete understanding of function composition. For $g(x) = \sqrt{x} + 1$, the domain is $x \ge 0$ and the range is $g(x) \ge 1$. This implies that the input to $f$ in $f(g(x))$ will always be greater than or equal to 1.

Step-by-Step Solution

We are given two real-valued functions $f(x)$ and $g(x)$ for $x \in \mathbb{R}$, with $g(x) = \sqrt{x} + 1$ and $f \circ g(x) = x + 3 - \sqrt{x}$. We need to find the value of $f(0)$.

Step 1: Express the components of $f \circ g(x)$ in terms of $g(x)$. The definition of $g(x)$ is $g(x) = \sqrt{x} + 1$. Our goal is to rewrite the expression for $f \circ g(x)$ in a way that clearly shows how $f$ acts on $g(x)$. To do this, we need to express $x$ and $\sqrt{x}$ in terms of $g(x)$.

From $g(x) = \sqrt{x} + 1$, we can isolate $\sqrt{x}$: $\sqrt{x} = g(x) - 1$ Reasoning: This allows us to substitute $\sqrt{x}$ in the expression for $f \circ g(x)$ with an expression solely in terms of $g(x)$.

Now, to express $x$ in terms of $g(x)$, we square both sides of the equation $\sqrt{x} = g(x) - 1$: $x = (\sqrt{x})^2 = (g(x) - 1)^2$ Expanding this, we get: $x = (g(x))^2 - 2g(x) + 1$ Reasoning: This step provides an expression for $x$ in terms of $g(x)$, enabling us to replace all occurrences of $x$ in the composite function's definition.

Step 2: Substitute the expressions for $x$ and $\sqrt{x}$ into $f \circ g(x)$. We are given $f \circ g(x) = x + 3 - \sqrt{x}$. Substitute $x = (g(x))^2 - 2g(x) + 1$ and $\sqrt{x} = g(x) - 1$ into the expression for $f \circ g(x)$: $f(g(x)) = [(g(x))^2 - 2g(x) + 1] + 3 - [g(x) - 1]$ Reasoning: By performing this substitution, we have successfully rewritten the right-hand side of the equation $f \circ g(x) = x + 3 - \sqrt{x}$ into an expression that depends only on $g(x)$. This prepares us to determine the general form of $f$.

Step 3: Simplify the expression for $f(g(x))$ to find the form of $f$. Combine the terms in the expression for $f(g(x))$: $f(g(x)) = (g(x))^2 - 2g(x) + 1 + 3 - g(x) + 1$ $f(g(x)) = (g(x))^2 + (-2g(x) - g(x)) + (1 + 3 + 1)$ $f(g(x)) = (g(x))^2 - 3g(x) + 5$ Reasoning: This simplified expression shows the relationship between the output of $g(x)$ and the output of $f$ when $g(x)$ is the input.

Step 4: Determine the general form of $f(y)$. Let $y = g(x)$. Then, the equation $f(g(x)) = (g(x))^2 - 3g(x) + 5$ becomes: $f(y) = y^2 - 3y + 5$ Reasoning: By replacing $g(x)$ with a generic variable $y$, we obtain the explicit formula for the function $f$. This form is valid for any value $y$ that can be an output of $g(x)$. Since the range of $g(x)$ is $[1, \infty)$, this form of $f(y)$ is derived for $y \ge 1$.

Step 5: Calculate $f(0)$. We need to find $f(0)$. Using the general form of $f(y) = y^2 - 3y + 5$, we substitute $y=0$: $f(0) = (0)^2 - 3(0) + 5$ $f(0) = 0 - 0 + 5$ $f(0) = 5$ Reasoning: Although the function $f(y)$ was derived for inputs $y \ge 1$ (the range of $g(x)$), the algebraic form $f(y) = y^2 - 3y + 5$ is a polynomial and is defined for all real numbers. Therefore, we can directly substitute $y=0$ into this polynomial to find $f(0)$. The problem asks for $f(0)$, not $f(g(x))$ where $g(x)=0$.

Common Mistakes & Tips

• Confusing $f(x)$ with $f(g(x))$: Ensure you are finding the explicit form of $f$ itself, not just manipulating the composite function's expression.
• Domain Issues: Be mindful of the domain of $g(x)$ and its implications for the inputs of $f$. In this case, the derived form of $f$ is valid for all real numbers, even though the inputs to $f$ in the composition were restricted.
• Algebraic Errors: Double-check your algebraic manipulations, especially when squaring expressions or combining like terms.

Summary

The problem required us to find $f(0)$ given $g(x) = \sqrt{x} + 1$ and $f \circ g(x) = x + 3 - \sqrt{x}$. The strategy involved expressing $x$ and $\sqrt{x}$ in terms of $g(x)$, substituting these into the composite function's definition, simplifying to find the general form of $f$, and then evaluating $f(0)$ using this form. By rewriting $f(g(x))$ as $(g(x))^2 - 3g(x) + 5$, we deduced that $f(y) = y^2 - 3y + 5$. Substituting $y=0$ into this expression yielded $f(0) = 5$.

The final answer is \boxed{5}.