Question

Let $f(x) = \dfrac{x}{2x-1}$ and $g(x) = \dfrac{x}{x-1}$. Then, the domain of the function \[ h(x) = f(g(x)) + g(f(x)) \] is all real numbers except:

• A. $\dfrac{1}{2},\, 1,\, \dfrac{3}{2}$
• B. $\dfrac{1}{2},\, 1$
• C. $-\dfrac{1}{2},\, \dfrac{1}{2},\, 1$
• D. $-1,\, \dfrac{1}{2},\, 1$

Hint:

When dealing with compositions of rational functions, 1.~Exclude values that make any denominator zero, and 2.~Also exclude values that make the inner function produce an invalid input for the outer function. Always combine all such restrictions at the end.

Answer 1

The Correct Option is D

Approach: Instead of hunting forbidden points abstractly, simplify each composite into a single clean fraction. The denominator of the simplified form, together with the inner restriction, exposes exactly which $x$-values are banned.

Step 1 — simplify $f(g(x))$: With $g(x) = \dfrac{x}{x-1}$, \[ f(g(x)) = \frac{g}{2g - 1} = \frac{\frac{x}{x-1}}{\frac{2x}{x-1} - 1} = \frac{x}{2x - (x-1)} = \frac{x}{x+1}. \] This simplified form bans $x = -1$; and the inner step used $x \ne 1$ (else $g$ itself is undefined). So $f(g(x))$ excludes $x = -1$ and $x = 1$.

Step 2 — simplify $g(f(x))$: With $f(x) = \dfrac{x}{2x-1}$, \[ g(f(x)) = \frac{f}{f - 1} = \frac{\frac{x}{2x-1}}{\frac{x}{2x-1} - 1} = \frac{x}{x - (2x-1)} = \frac{x}{1 - x}. \] This bans $x = 1$; and the inner step used $x \ne \tfrac12$ (else $f$ is undefined). So $g(f(x))$ excludes $x = \tfrac12$ and $x = 1$.

Step 3 — union of all bans: $\{-1,\ 1\} \cup \{\tfrac12,\ 1\} = \{-1,\ \tfrac12,\ 1\}$.

Final answer: All reals except $-1,\ \dfrac12,\ 1$ — option 4.