Question

In the graph of the following piecewise function, which coordinate point will have an open circle?

f(x)={2x+2 if x>4
1/2x+1 if x≤2

Answer 1

wrong

Answer 2

To find the coordinate point with an open circle in the given piecewise function, let's analyze the function:

\( f(x) = \begin{cases}
2x + 2 & \text{if } x > 4 \\
\frac{1}{2}x + 1 & \text{if } x \leq 2
\end{cases} \)

1. **Condition for \( x > 4 \)**: For \( x \) values greater than 4, the function is given by \( f(x) = 2x + 2 \). Since this function applies to all values greater than 4, there is no open circle at any point in this range because the function is defined there.

2. **Condition for \( x \leq 2 \)**: For \( x \) values less than or equal to 2, the function is given by \( f(x) = \frac{1}{2}x + 1 \). Since this is defined for \( x \) values less than or equal to 2, there will be no open circle in this portion either, except at \( x = 2 \).

Now, we need to check what happens at \( x = 2 \):
- When \( x = 2 \), we can substitute it into the function defined for that interval:
\[
f(2) = \frac{1}{2}(2) + 1 = 1 + 1 = 2
\]
- However, since the second piece of the function (i.e., \( 2x + 2 \)) does not include \( x = 2 \) (it is defined only for \( x > 4 \)), there will be an open circle at the point \( (2, 2) \).

Thus, the coordinate point with an open circle in the graph of this piecewise function is:
\[
(2, 2)
\]