Let's analyze the given piecewise function to determine if there are any points of discontinuity. The function is defined as follows:
[tex]\[
f(x)=
\begin{cases}
-(x+1)^2 + 2 & \text{for } -\infty < x < -1 \\
-x + 2 & \text{for } -1 \leq x < 2 \\
\sqrt{x-1} & \text{for } 2 \leq x < \infty
\end{cases}
\][/tex]
To find points of discontinuity, we will examine the transition points [tex]\( x = -1 \)[/tex] and [tex]\( x = 2 \)[/tex]. Here’s how:
### Step 1: Check the point [tex]\( x = -1 \)[/tex]
We need to ensure that the left-hand limit (as [tex]\( x \)[/tex] approaches -1 from the left) equals the right-hand limit (as [tex]\( x \)[/tex] approaches -1 from the right), and that both are equal to the value of the function at [tex]\( x = -1 \)[/tex].
- Left-hand limit at [tex]\( x = -1 \)[/tex]:
For [tex]\( x < -1 \)[/tex], [tex]\( f(x) = -(x+1)^2 + 2 \)[/tex]:
[tex]\[
\lim_{x \to -1^-} f(x) = \lim_{x \to -1^-} [-(x+1)^2 + 2] = -( (-1+1)^2 ) + 2 = 2
\][/tex]
- Right-hand limit at [tex]\( x = -1 \)[/tex]:
For [tex]\( -1 \leq x < 2 \)[/tex], [tex]\( f(x) = -x + 2 \)[/tex]:
[tex]\[
\lim_{x \to -1^+} f(x) = \lim_{x \to -1^+} [-x + 2] = -(-1) + 2 = 3
\][/tex]
Since the left-hand limit (2) does not equal the right-hand limit (3), there is a discontinuity at [tex]\( x = -1 \)[/tex].
### Step 2: Check the point [tex]\( x = 2 \)[/tex]
We again check both left-hand and right-hand limits at [tex]\( x \)[/tex].
- Left-hand limit at [tex]\( x = 2 \)[/tex]:
For [tex]\( -1 \leq x < 2 \)[/tex], [tex]\( f(x) = -x + 2 \)[/tex]:
[tex]\[
\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} [-x + 2] = -(2) + 2 = 0
\][/tex]
- Right-hand limit at [tex]\( x = 2 \)[/tex]:
For [tex]\( 2 \leq x < \infty \)[/tex], [tex]\( f(x) = \sqrt{x-1} \)[/tex]:
[tex]\[
\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} \sqrt{x-1} = \sqrt{2-1} = 1
\][/tex]
Since the left-hand limit (0) does not equal the right-hand limit (1), there is a discontinuity at [tex]\( x = 2 \)[/tex].
### Conclusion:
We examined crucial points and found that neither [tex]\( x = -1 \)[/tex] nor [tex]\( x = 2 \)[/tex] have continuous values at the boundary points. Therefore, there are no points of discontinuity in the piecewise function [tex]\( f(x) \)[/tex].
The correct answer is:
[tex]\[ \boxed{\text{D. no points of discontinuity}} \][/tex]
