To determine where the function [tex]\( f(x) \)[/tex] has a discontinuity, we need to examine the points where the definition of the function changes and match the left-hand limit and the right-hand limit at those points.
The function given is:
[tex]\[ f(x) =
\begin{cases}
x + 1 & \text{if } x < -2 \\
2x - 3 & \text{if } x \geq -2
\end{cases}
\][/tex]
Let's check the point where the definition changes, which is at [tex]\( x = -2 \)[/tex].
### Left-hand limit:
For the part of the function where [tex]\( x < -2 \)[/tex]:
[tex]\[ \lim_{{x \to -2^-}} f(x) = \lim_{{x \to -2^-}} (x + 1) \][/tex]
Substituting [tex]\( x = -2 \)[/tex]:
[tex]\[ \lim_{{x \to -2^-}} (x + 1) = -2 + 1 = -1 \][/tex]
### Right-hand limit:
For the part of the function where [tex]\( x \geq -2 \)[/tex]:
[tex]\[ \lim_{{x \to -2^+}} f(x) = \lim_{{x \to -2^+}} (2x - 3) \][/tex]
Substituting [tex]\( x = -2 \)[/tex]:
[tex]\[ \lim_{{x \to -2^+}} (2x - 3) = 2(-2) - 3 = -4 - 3 = -7 \][/tex]
### Comparison of limits:
For the function to be continuous at [tex]\( x = -2 \)[/tex], the left-hand limit and the right-hand limit must be equal.
In this case:
[tex]\[ \lim_{{x \to -2^-}} f(x) = -1 \][/tex]
[tex]\[ \lim_{{x \to -2^+}} f(x) = -7 \][/tex]
Since [tex]\(-1 \neq -7\)[/tex], the function is not continuous at [tex]\( x = -2 \)[/tex].
### Conclusion:
Therefore, the function [tex]\( f(x) \)[/tex] has a discontinuity at [tex]\( x = -2 \)[/tex].
