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: Recall the meaning of composition.
$(f\circ g)(x)=f\bigl(g(x)\bigr)$, so we feed $g(x)=\dfrac{x}{x+1}$ into $f(t)=\dfrac{t}{2t+1}$.
Step 2: Use a clean substitution variable.
Let $t=\dfrac{x}{x+1}$. Then $f(t)=\dfrac{t}{2t+1}$.
Step 3: Compute the numerator and denominator separately.
Numerator: $t=\dfrac{x}{x+1}$. Denominator: $2t+1=\dfrac{2x}{x+1}+1=\dfrac{2x+(x+1)}{x+1}=\dfrac{3x+1}{x+1}$.
Step 4: Divide the two fractions.
\[ f(t)=\frac{\dfrac{x}{x+1}}{\dfrac{3x+1}{x+1}}=\frac{x}{x+1}\cdot\frac{x+1}{3x+1}. \]
Step 5: Cancel the common factor.
The $(x+1)$ terms cancel, leaving $\dfrac{x}{3x+1}$.
Step 6: Sanity check with a number.
At $x=1$: $g(1)=\tfrac12$, $f(\tfrac12)=\tfrac{1/2}{2}= \tfrac14$, and $\tfrac{1}{3+1}=\tfrac14$ agrees. \[ \boxed{(f\circ g)(x)=\dfrac{x}{3x+1}} \]