To graph the piecewise function ( f(x) = \begin{cases} x + 5 & \text{if } x \leq -2 \ 2x + 3 & \text{if } x > -2 \end{cases} ), we need to analyze the two pieces separately.
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For the first piece, \( f(x) = x + 5 \) when \( x \leq -2 \):
- At \( x = -2 \), \( f(-2) = -2 + 5 = 3 \). This point will be included in the graph as the range allows \( x \) to take the value \(-2\).
- The graph is a line (slope of 1) that extends to the left from the point \((-2, 3)\).
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For the second piece, \( f(x) = 2x + 3 \) when \( x > -2 \):
- At \( x = -2 \), \( f(-2) \) is not included here because this piece is defined for \( x > -2\). Instead, we can find the limit:
- \( \lim_{x \to -2^+} f(x) = 2(-2) + 3 = -1 \). This means that as \( x \) approaches \(-2\) from the right, the function approaches \(-1\).
- The graph for this piece starts just to the right of \((-2, -1)\) and continues to the right.
Now, looking for open or closed endpoints:
- At \( x = -2 \), the point \((-2, 3)\) is closed (included) and the point \((-2, -1)\) is open (not included).
Thus, there is an open endpoint at \((-2, -1)\).
So the correct answer is: Yes, at (−2, −1).