If `f(x)=(x^(2)-9)/(x^(2)-2x-3),xne3` is continuous at x = 3, then which one of the following is correct?

Options

A. A) `f(3)=0`
B. B) `f(3)=1.5`
C. C) `f(3)=3`
D. D) `f(3)=-1.5`

Answer

To determine the correct value of the function \( f(x) = \frac{x^2 - 9}{x^2 - 2x - 3} \) at \( x = 3 \) and ensure it is continuous at that point, we will follow these steps:

### Step 1: Understand the continuity condition
A function is continuous at a point \( x = a \) if:
\[ f(a) = \lim_{x \to a} f(x) \]
In this case, we need to find \( f(3) \) and check if it equals \( \lim_{x \to 3} f(x) \).

### Step 2: Calculate the limit as \( x \) approaches 3
We will compute:
\[ \lim_{x \to 3} f(x) = \lim_{x \to 3} \frac{x^2 - 9}{x^2 - 2x - 3} \]

### Step 3: Factor the numerator and denominator
The numerator \( x^2 - 9 \) can be factored as:
\[ x^2 - 9 = (x - 3)(x + 3) \]
The denominator \( x^2 - 2x - 3 \) can be factored as:
\[ x^2 - 2x - 3 = (x - 3)(x + 1) \]

### Step 4: Rewrite the function
Now we can rewrite \( f(x) \):
\[ f(x) = \frac{(x - 3)(x + 3)}{(x - 3)(x + 1)} \]
Since \( x \neq 3 \), we can cancel \( (x - 3) \) from the numerator and denominator:
\[ f(x) = \frac{x + 3}{x + 1} \quad \text{for } x \neq 3 \]

### Step 5: Calculate the limit
Now we compute the limit:
\[ \lim_{x \to 3} f(x) = \lim_{x \to 3} \frac{x + 3}{x + 1} \]
Substituting \( x = 3 \):
\[ = \frac{3 + 3}{3 + 1} = \frac{6}{4} = \frac{3}{2} = 1.5 \]

### Step 6: Conclusion
Thus, we find that:
\[ \lim_{x \to 3} f(x) = 1.5 \]
For \( f(x) \) to be continuous at \( x = 3 \), we must define \( f(3) = 1.5 \).

### Final Answer
The correct option is that \( f(3) = 1.5 \).

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