To determine if the function [tex]\( f(x) = \frac{x^2 + 4x + 3}{x+1} \)[/tex] has a discontinuity at [tex]\( x = -1 \)[/tex], we need to analyze its behavior as [tex]\( x \)[/tex] approaches -1 from both sides (left and right).
1. Simplify the Function:
First, let's simplify the given function to identify any possible points of discontinuity.
[tex]\[
f(x) = \frac{x^2 + 4x + 3}{x + 1}
\][/tex]
Factor the numerator:
[tex]\[
x^2 + 4x + 3 = (x + 1)(x + 3)
\][/tex]
So, the function becomes:
[tex]\[
f(x) = \frac{(x + 1)(x + 3)}{x + 1}
\][/tex]
For [tex]\( x \neq -1 \)[/tex], we can cancel out [tex]\( (x + 1) \)[/tex]:
[tex]\[
f(x) = x + 3 \quad \text{when} \quad x \neq -1
\][/tex]
Therefore, the simplified function is [tex]\( f(x) = x + 3 \)[/tex] for all [tex]\( x \neq -1 \)[/tex].
2. Check the Limit from the Left:
We now check the limit of [tex]\( f(x) \)[/tex] as [tex]\( x \)[/tex] approaches -1 from the left ([tex]\( x \to -1^- \)[/tex]):
[tex]\[
\lim_{x \to -1^-} f(x) = \lim_{x \to -1^-} (x + 3) = -1 + 3 = 2
\][/tex]
3. Check the Limit from the Right:
Similarly, we check the limit of [tex]\( f(x) \)[/tex] as [tex]\( x \)[/tex] approaches -1 from the right ([tex]\( x \to -1^+ \)[/tex]):
[tex]\[
\lim_{x \to -1^+} f(x) = \lim_{x \to -1^+} (x + 3) = -1 + 3 = 2
\][/tex]
4. Comparison of Limits:
Since the limit from the left ([tex]\( 2 \)[/tex]) is equal to the limit from the right ([tex]\( 2 \)[/tex]), we have:
[tex]\[
\lim_{x \to -1^-} f(x) = \lim_{x \to -1^+} f(x) = 2
\][/tex]
5. Determine Continuity:
Now, we need to check the value of the function at [tex]\( x = -1 \)[/tex]. Since the original function [tex]\( f(x) = \frac{x^2 + 4x + 3}{x + 1} \)[/tex] is undefined at [tex]\( x = -1 \)[/tex] (as the denominator becomes zero), we cannot assign a function value to [tex]\( f(-1) \)[/tex].
However, the essential part for determining a discontinuity at a point is to check if the left-hand limit and right-hand limit are equal and if the function is defined there. Although [tex]\( f(-1) \)[/tex] is not defined in the original function, both limits to [tex]\( -1 \)[/tex] are equal and finite.
6. Conclusion:
Despite [tex]\( f(x) \)[/tex] being undefined at [tex]\( x = -1 \)[/tex], the limits from both sides approaching [tex]\( -1 \)[/tex] are equal.
Therefore, there is no discontinuity in the sense that the limits align smoothly at [tex]\( x = -1 \)[/tex], but technically, the function [tex]\( f(x) \)[/tex] is not defined at [tex]\( x = -1 \)[/tex].
Considering all points above:
The correct term would be that [tex]\( f(x) \)[/tex] is not continuous at [tex]\( x = -1 \)[/tex] due to it not having a defined value there.
So the answer to "Does this function have a discontinuity at [tex]\( x = -1 \)[/tex]?" is: No in the sense that both limits from either side are continuous and meet at the same value, so no jump or infinite discontinuity exists there.
