Question

If $f(x) = \frac{x}{2x+1}$ and $g(x) = \frac{x}{x+1}$, then $(f \circ g)(x) = $

• A. $\frac{2x-1}{x+1}$
• B. $\frac{x}{3x+1}$
• C. $\frac{x+1}{x+2}$
• D. $\frac{x-1}{2x+1}$

Hint:

To verify functional composition choices in seconds during an exam, test a small value like $x = 1$. Here, $g(1) = \frac{1}{1+1} = \frac{1}{2}$. Then compute $f(\frac{1}{2}) = \frac{1/2}{2(1/2)+1} = \frac{1/2}{2} = \frac{1}{4}$. Plug $x=1$ into the choices to see which one equals $\frac{1}{4}$. Option (B) gives $\frac{1}{3(1)+1} = \frac{1}{4}$, confirming it instantly!

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The problem requires computing the composite function $(f \circ g)(x)$, which means evaluating the function $f(x)$ using $g(x)$ as its input variable.

Step 2: Key Formula or Approach:
The definition of a composite function tells us that: $$(f \circ g)(x) = f(g(x))$$ We will take the algebraic expression for $g(x)$ and substitute it wherever an $x$ appears inside the formula for $f(x)$.

Step 3: Detailed Explanation:
We are given: $$f(x) = \frac{x}{2x+1} \quad \text{and} \quad g(x) = \frac{x}{x+1}$$ Substitute $g(x)$ into the function $f$: $$f(g(x)) = \frac{g(x)}{2g(x) + 1} = \frac{\frac{x}{x+1}}{2\left(\frac{x}{x+1}\right) + 1}$$ To simplify the denominator expression, find a common denominator: $$2\left(\frac{x}{x+1}\right) + 1 = \frac{2x}{x+1} + \frac{x+1}{x+1} = \frac{2x + x + 1}{x+1} = \frac{3x+1}{x+1}$$ Now, substitute this simplified denominator back into our fraction compound: $$f(g(x)) = \frac{\frac{x}{x+1}}{\frac{3x+1}{x+1}}$$ Cancel the common $(x+1)$ term from both denominators: $$f(g(x)) = \frac{x}{3x+1}$$

Step 4: Final Answer:
The composite function resolves to $\frac{x}{3x+1}$, which matches option (B).