The function `f(x)=(x^2-1)/(x^3-1)` is not defined for x=1. The value of f(1) so that the function extended by this value is continuous is

Options

A. `2/3`
B. `1/3`
C. 1
D. 0

Answer

To find the value of \( f(1) \) such that the function \( f(x) = \frac{x^2 - 1}{x^3 - 1} \) is continuous at \( x = 1 \), we need to follow these steps:

### Step 1: Identify the point of discontinuity
The function \( f(x) \) is not defined at \( x = 1 \) because both the numerator and denominator become zero at this point:
\[ f(1) = \frac{1^2 - 1}{1^3 - 1} = \frac{0}{0} \]
This indicates a removable discontinuity.

### Step 2: Factor the numerator and denominator
We can factor both the numerator and the denominator:
- The numerator \( x^2 - 1 \) can be factored as:
\[ x^2 - 1 = (x - 1)(x + 1) \]
- The denominator \( x^3 - 1 \) can be factored using the difference of cubes:
\[ x^3 - 1 = (x - 1)(x^2 + x + 1) \]

### Step 3: Simplify the function
Substituting the factored forms into the function gives:
\[ f(x) = \frac{(x - 1)(x + 1)}{(x - 1)(x^2 + x + 1)} \]
For \( x \neq 1 \), we can cancel \( (x - 1) \):
\[ f(x) = \frac{x + 1}{x^2 + x + 1} \]

### Step 4: Find the limit as \( x \) approaches 1
To find the value of \( f(1) \) that makes the function continuous, we need to calculate:
\[ \lim_{x \to 1} f(x) = \lim_{x \to 1} \frac{x + 1}{x^2 + x + 1} \]
Substituting \( x = 1 \) into the simplified function:
\[ \lim_{x \to 1} f(x) = \frac{1 + 1}{1^2 + 1 + 1} = \frac{2}{3} \]

### Step 5: Define \( f(1) \)
For the function to be continuous at \( x = 1 \), we set:
\[ f(1) = \frac{2}{3} \]

Thus, the value of \( f(1) \) that makes the function continuous is \( \frac{2}{3} \).

### Final Answer:
\[ f(1) = \frac{2}{3} \]

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