Asked by hi
In the graph of the following piecewise function, which coordinate point will have an open circle?
f(x)={2x+2 if x>412x+1 if x≤2
f(x)={2x+2 if x>412x+1 if x≤2
Answers
Answered by
GPT-4o mini
To determine where the open circle will be located for the piecewise function given by:
\[
f(x) = \begin{cases}
2x + 2 & \text{if } x > 2 \\
2x + 1 & \text{if } x \leq 2
\end{cases}
\]
we need to evaluate the function at the point where the definition of the function changes, which is at \(x = 2\).
1. **Evaluate \( f(2) \)** using the second piece (since it is defined for \(x \leq 2\)):
\[
f(2) = 2(2) + 1 = 4 + 1 = 5
\]
So, when \(x = 2\), \(f(2) = 5\).
2. **For \(x > 2\)**, we switch to the first piece of the function. We can check the limit as \(x\) approaches 2 from the right:
\[
\lim_{x \to 2^+} f(x) = 2(2) + 2 = 4 + 2 = 6
\]
Thus:
- At \(x = 2\), the function \(f(x)\) is defined and equals 5, indicating a filled circle at the point \((2, 5)\).
- Approaching from the right, the function's value approaches 6, which means there will be an open circle at \((2, 6)\) because the value of the function for \(x > 2\) is not defined at \(x = 2\) (it takes the value of 5 from the left, indicated by the closed circle, while approaching the value 6 from the right).
Therefore, the coordinate point that will have an open circle is:
\[
\boxed{(2, 6)}
\]
\[
f(x) = \begin{cases}
2x + 2 & \text{if } x > 2 \\
2x + 1 & \text{if } x \leq 2
\end{cases}
\]
we need to evaluate the function at the point where the definition of the function changes, which is at \(x = 2\).
1. **Evaluate \( f(2) \)** using the second piece (since it is defined for \(x \leq 2\)):
\[
f(2) = 2(2) + 1 = 4 + 1 = 5
\]
So, when \(x = 2\), \(f(2) = 5\).
2. **For \(x > 2\)**, we switch to the first piece of the function. We can check the limit as \(x\) approaches 2 from the right:
\[
\lim_{x \to 2^+} f(x) = 2(2) + 2 = 4 + 2 = 6
\]
Thus:
- At \(x = 2\), the function \(f(x)\) is defined and equals 5, indicating a filled circle at the point \((2, 5)\).
- Approaching from the right, the function's value approaches 6, which means there will be an open circle at \((2, 6)\) because the value of the function for \(x > 2\) is not defined at \(x = 2\) (it takes the value of 5 from the left, indicated by the closed circle, while approaching the value 6 from the right).
Therefore, the coordinate point that will have an open circle is:
\[
\boxed{(2, 6)}
\]