To determine the points at which the given piecewise function [tex]\( f(x) \)[/tex] is discontinuous, we need to check the continuity at the boundaries of the piecewise intervals.
The function [tex]\( f(x) \)[/tex] is defined as follows:
[tex]\[
f(x) =
\begin{cases}
x + 2 & \text{if } x \leq 0 \\
2 & \text{if } 0 < x \leq 1 \\
x^2 + 2 & \text{if } x > 1
\end{cases}
\][/tex]
The boundaries of the intervals are at [tex]\( x = 0 \)[/tex] and [tex]\( x = 1 \)[/tex]. We'll check for continuity at these points by verifying if the left-hand limit, right-hand limit, and the function value at these points match.
### Check at [tex]\( x = 0 \)[/tex]:
1. Left-hand limit as [tex]\( x \to 0^- \)[/tex]:
[tex]\[
\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (x + 2) = 0 + 2 = 2
\][/tex]
2. Right-hand limit as [tex]\( x \to 0^+ \)[/tex]:
[tex]\[
\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} 2 = 2
\][/tex]
3. Function value at [tex]\( x = 0 \)[/tex]:
[tex]\[
f(0) = 0 + 2 = 2
\][/tex]
Since the left-hand limit, right-hand limit, and the function value at [tex]\( x = 0 \)[/tex] are all equal to 2, the function [tex]\( f(x) \)[/tex] is continuous at [tex]\( x = 0 \)[/tex].
### Check at [tex]\( x = 1 \)[/tex]:
1. Left-hand limit as [tex]\( x \to 1^- \)[/tex]:
[tex]\[
\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} 2 = 2
\][/tex]
2. Right-hand limit as [tex]\( x \to 1^+ \)[/tex]:
[tex]\[
\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (x^2 + 2) = 1^2 + 2 = 1 + 2 = 3
\][/tex]
3. Function value at [tex]\( x = 1 \)[/tex]:
[tex]\[
f(1) = 2
\][/tex]
At [tex]\( x = 1 \)[/tex], the left-hand limit is 2, while the right-hand limit is 3. Since these limits are not equal, the function [tex]\( f(x) \)[/tex] is discontinuous at [tex]\( x = 1 \)[/tex].
### Conclusion:
The function [tex]\( f(x) \)[/tex] is discontinuous at [tex]\( x = 1 \)[/tex]. This is the only point among the boundaries of the piecewise function intervals where the function exhibits discontinuity.
