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

2 answers

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: \[ (2, 2) \]

wrong