To determine the open endpoints in the piecewise function \( f(x) \), we start by analyzing the function:
\[ f(x) = \begin{cases} x + 5 & \text{if } x \leq -2 \ 2x + 3 & \text{if } x > -2 \end{cases} \]
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For \( x \leq -2 \): \[ f(x) = x + 5 \] At \( x = -2 \): \[ f(-2) = -2 + 5 = 3 \] This means the point \( (-2, 3) \) is included in the graph since the condition \( x \leq -2 \) is met.
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For \( x > -2 \): \[ f(x) = 2x + 3 \] However, when \( x = -2 \), this piece does not apply (since \( -2 \) is not greater than \( -2 \)). The value of the function approaches \( 3 \) as \( x \) approaches \( -2 \) from the right (but never quite reaches it).
Thus, we can see that there should be an open endpoint at \( (-2, 3) \) because there is no value of \( f(x) \) defined for \( x = -2 \) in the second piece of the function.
- Checking continuity at \( x = -2 \): \[ \lim_{x \to -2^-} f(x) = 3 \quad (\text{using } x + 5) \] \[ \lim_{x \to -2^+} f(x) = 3 \quad (\text{using } 2x + 3) \] The limit from both sides equals \( 3 \), but \( f(-2) = 3 \) from the first piece means we can include the point but leave the endpoint open for the function definition on the right.
Therefore, the graph will indeed have an open endpoint at \( (-2, 3) \) (not \( (-2, -1) \) which is not part of the function).
Final answer:
Yes, at \( (-2, 3) \).