If `f(x) = (x + 1)/(x-1) and g(x) = (1)/(x-2)`, then (fog)(x) is discontinuous at

Options

A. (a) x = 3 only
B. (b) x = 2 only
C. (c) x = 2 and 3 only
D. (d) x = 1 only

Answer

To solve the problem, we need to find the composition of the functions \( f(x) \) and \( g(x) \), denoted as \( (f \circ g)(x) \), and determine where this function is discontinuous.

### Step 1: Define the functions
We have:
\[ f(x) = \frac{x + 1}{x - 1} \]
\[ g(x) = \frac{1}{x - 2} \]

### Step 2: Find \( (f \circ g)(x) \)
To find \( (f \circ g)(x) \), we need to substitute \( g(x) \) into \( f(x) \):
\[ (f \circ g)(x) = f(g(x)) = f\left(\frac{1}{x - 2}\right) \]

### Step 3: Substitute \( g(x) \) into \( f(x) \)
Now, we substitute \( g(x) \) into \( f(x) \):
\[ f\left(\frac{1}{x - 2}\right) = \frac{\frac{1}{x - 2} + 1}{\frac{1}{x - 2} - 1} \]

### Step 4: Simplify the expression
We simplify the numerator and denominator:
- **Numerator**:
\[ \frac{1}{x - 2} + 1 = \frac{1 + (x - 2)}{x - 2} = \frac{x - 1}{x - 2} \]

- **Denominator**:
\[ \frac{1}{x - 2} - 1 = \frac{1 - (x - 2)}{x - 2} = \frac{3 - x}{x - 2} \]

Putting it all together:
\[ f\left(\frac{1}{x - 2}\right) = \frac{\frac{x - 1}{x - 2}}{\frac{3 - x}{x - 2}} = \frac{x - 1}{3 - x} \]

### Step 5: Identify points of discontinuity
The function \( f(g(x)) = \frac{x - 1}{3 - x} \) is discontinuous where the denominator is zero. Therefore, we set:
\[ 3 - x = 0 \implies x = 3 \]

### Conclusion
Thus, \( (f \circ g)(x) \) is discontinuous at \( x = 3 \).

### Final Answer
The function \( (f \circ g)(x) \) is discontinuous at \( x = 3 \).

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