Graph the piecewise function f(x)={x+5 if x≤−22x+3 if x>−2. Will the graph have any open endpoints? If yes, where?(1 point)

Question

Responses

Yes, at (−2,3) and (−2,−1).
Yes, at left parenthesis negative 2 comma 3 right parenthesis and left parenthesis negative 2 comma negative 1 right parenthesis .

Yes, at (−2,−1).
Yes, at left parenthesis negative 2 comma negative 1 right parenthesis .

Yes, at (−2,3).
Yes, at left parenthesis negative 2 comma 3 right parenthesis .

No, all endpoints are closed.

Answer 1

To analyze the piecewise function \( f(x) \) defined as:

\[ f(x) = \begin{cases} x + 5 & \text{if } x \leq -2 \ 2x + 3 & \text{if } x > -2 \end{cases} \]

we need to determine the behavior of the function at the boundary point \( x = -2 \).

1. For \( x \leq -2 \): Use the equation \( f(x) = x + 5 \).

   • When \( x = -2 \): \[ f(-2) = -2 + 5 = 3 \] So, the function has a closed endpoint at the point \((-2, 3)\).

2. For \( x > -2 \): Use the equation \( f(x) = 2x + 3 \).

   • As \( x \) approaches \(-2\) from the right (greater than -2): \[ f(-2) = 2(-2) + 3 = -4 + 3 = -1 \] Here, the function approaches \((-2, -1)\) but does not actually reach that point since the domain is restricted to \( x > -2\). Thus, there is an open endpoint at this point.

In conclusion, the graph will have:

• A closed endpoint at \((-2, 3)\) because \( x = -2 \) is included in the first piece of the function.
• An open endpoint at \((-2, -1)\) because \( x = -2 \) is not included in the second piece of the function.

The correct response is:

Yes, at (−2, 3) and (−2, −1).