If `f(x)` is continuous at `x=3`, where `f(x)=(x^(2)-7x+12)/(x^(2)-5x+6)`,for `x!=3`, then `f(3)=`

Options

A. -1
B. 1
C. `1/5`
D. `7/5`

Answer

To find \( f(3) \) for the function

\[ f(x) = \frac{x^2 - 7x + 12}{x^2 - 5x + 6} \]

which is continuous at \( x = 3 \), we first need to simplify the function.

### Step 1: Factor the numerator and denominator

The numerator \( x^2 - 7x + 12 \) can be factored as follows:

\[ x^2 - 7x + 12 = (x - 3)(x - 4) \]

The denominator \( x^2 - 5x + 6 \) can be factored as:

\[ x^2 - 5x + 6 = (x - 2)(x - 3) \]

### Step 2: Rewrite the function

Now we can rewrite \( f(x) \):

\[ f(x) = \frac{(x - 3)(x - 4)}{(x - 2)(x - 3)} \]

### Step 3: Cancel the common factor

Since \( f(x) \) is defined for \( x \neq 3 \), we can cancel the \( (x - 3) \) term:

\[ f(x) = \frac{x - 4}{x - 2} \quad \text{for } x \neq 3 \]

### Step 4: Find the limit as \( x \) approaches 3

To find \( f(3) \), we need to evaluate the limit of \( f(x) \) as \( x \) approaches 3:

\[ \lim_{x \to 3} f(x) = \lim_{x \to 3} \frac{x - 4}{x - 2} \]

Substituting \( x = 3 \):

\[ \lim_{x \to 3} f(x) = \frac{3 - 4}{3 - 2} = \frac{-1}{1} = -1 \]

### Step 5: Conclusion

Since \( f(x) \) is continuous at \( x = 3 \), we have:

\[ f(3) = \lim_{x \to 3} f(x) = -1 \]

Thus, the value of \( f(3) \) is:

\[ \boxed{-1} \]