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

To determine the domain of the function \(h(x) = f(g(x)) + g(f(x))\), where \(f(x) = \dfrac{x}{2x-1}\) and \(g(x) = \dfrac{x}{x-1}\), we need to find where each function is defined and check for any common restrictions.

Step 1: Determine the domain of \(g(x)\)

The function \(g(x) = \dfrac{x}{x-1}\) is defined for all real numbers except where the denominator is zero:

• Set the denominator equal to zero: \(x-1 = 0 \Rightarrow x = 1\)

Thus, the domain of \(g(x)\) is all real numbers except \(1\).

Step 2: Determine the domain of \(f(x)\)

The function \(f(x) = \dfrac{x}{2x-1}\) is defined for all real numbers except where the denominator is zero:

• Set the denominator equal to zero: \(2x-1 = 0 \Rightarrow x = \dfrac{1}{2}\)

Thus, the domain of \(f(x)\) is all real numbers except \(\dfrac{1}{2}\).

Step 3: Determine the domain of \(f(g(x)) = f\left(\dfrac{x}{x-1}\right)\)

First, check where \(g(x)\) is defined, i.e., \(x \neq 1\). Then determine when \(f(y)\) is defined where \(y = g(x) = \dfrac{x}{x-1}\):

• Set \(2y-1 = 0\):
  \(2\left(\dfrac{x}{x-1}\right)-1 = 0 \Rightarrow \dfrac{2x}{x-1} = 1 \Rightarrow 2x = x-1 \Rightarrow x = -1\)
• Thus, \(f(g(x))\) is undefined when \(x = -1\) and \(x = 1\).

Step 4: Determine the domain of \(g(f(x)) = g\left(\dfrac{x}{2x-1}\right)\)

First, check where \(f(x)\) is defined, i.e., \(x \neq \dfrac{1}{2}\). Then determine when \(g(y)\) is defined where \(y = f(x) = \dfrac{x}{2x-1}\):

• Set \(y-1 = 0\):
  \(\dfrac{x}{2x-1} - 1 = 0 \Rightarrow \dfrac{x-(2x-1)}{2x-1} = 0 \Rightarrow -x+1 = 0 \Rightarrow x = 1\)
• Thus, \(g(f(x))\) is undefined when \(x = \dfrac{1}{2}\) and \(x = 1\).

Conclusion:

The function \(h(x) = f(g(x)) + g(f(x))\) is undefined at \(x = -1\), \(x = \dfrac{1}{2}\) and \(x = 1\). Thus, the domain of \(h(x)\) is all real numbers except \(-1\), \(\dfrac{1}{2}\), and \(1\).

Therefore, the correct answer is: \(-1,\, \dfrac{1}{2},\, 1\).

Answer 2

To determine the domain of \[ h(x) = f(g(x)) + g(f(x)), \] both expressions \(f(g(x))\) and \(g(f(x))\) must be defined. 1. Domain of \(f(g(x))\) Inner function: \[ g(x) = \frac{x}{x-1} \] is undefined when the denominator is zero: \[ x-1 \neq 0 \quad \Rightarrow \quad x \neq 1. \] Outer function: \[ f(u) = \frac{u}{2u-1} \] is undefined when \[ 2u - 1 = 0 \quad \Rightarrow \quad u = \frac{1}{2}. \] Thus, we need \[ g(x) \neq \frac{1}{2}. \] Solve: \[ \frac{x}{x-1} \neq \frac12 \] Cross–multiply: \[ 2x \neq x - 1 \quad\Rightarrow\quad x \neq -1. \] Hence, for \(f(g(x))\), the restrictions are: \[ x \neq 1,\; -1. \] 2. Domain of \(g(f(x))\) Inner function: \[ f(x) = \frac{x}{2x-1} \] is undefined when \[ 2x - 1 = 0 \quad \Rightarrow\quad x \neq \frac12. \] Outer function: \[ g(v) = \frac{v}{v-1} \] is undefined when \(v = 1\). So we need: \[ f(x) \neq 1. \] Solve: \[ \frac{x}{2x-1} \neq 1. \] Set inequality: \[ x \neq 2x - 1 \quad\Rightarrow\quad -x \neq -1 \quad\Rightarrow\quad x \neq 1. \] Thus, for \(g(f(x))\): \[ x \neq \frac12,\; 1. \] 3. Combine all restrictions From \(f(g(x))\): \(x \neq 1,\; -1\). From \(g(f(x))\): \(x \neq \frac12,\; 1\). Therefore, the domain excludes: \[ \boxed{-1,\;\frac12,\;1}. \]