To determine the behavior of the function [tex]\( f(x) = \frac{x-3}{x^2-4} \)[/tex] at [tex]\( x = 2 \)[/tex], let's analyze the function step-by-step.
1. Identify the Denominator and the Numerator:
- The numerator of the function is [tex]\( x - 3 \)[/tex].
- The denominator of the function is [tex]\( x^2 - 4 \)[/tex].
2. Factorize the Denominator:
- We can factorize the denominator [tex]\( x^2 - 4 \)[/tex] using the difference of squares:
[tex]\[
x^2 - 4 = (x - 2)(x + 2)
\][/tex]
3. Check the Value of the Denominator at [tex]\( x = 2 \)[/tex]:
- Substitute [tex]\( x = 2 \)[/tex] into the denominator:
[tex]\[
x^2 - 4 = (2 - 2)(2 + 2) = 0
\][/tex]
4. Behavior of the Function at [tex]\( x = 2 \)[/tex]:
- The denominator becomes zero when [tex]\( x = 2 \)[/tex], which means the function is undefined at this point. This can lead to a discontinuity.
- Next, check the type of this discontinuity. Observe the form of the numerator at [tex]\( x = 2 \)[/tex]:
[tex]\[
x - 3 = 2 - 3 = -1
\][/tex]
- Here, the numerator does not become zero, so the fraction [tex]\(\frac{-1}{(x-2)(x+2)}\)[/tex] leads to [tex]\(\frac{-1}{0}\)[/tex], indicating an undefined and infinite behavior.
5. Conclusion:
- Since the denominator equals zero and the numerator does not equal zero at [tex]\( x = 2 \)[/tex], [tex]\( f(x) \)[/tex] has a non-removable discontinuity, specifically an infinite discontinuity, at this point.
Therefore, the function [tex]\( f(x) = \frac{x-3}{x^2-4} \)[/tex] is discontinuous with an infinite discontinuity at [tex]\( x = 2 \)[/tex].
The correct answer is:
- discontinuous with an infinite discontinuity.
