Question

explain how to graph the given piece - wise - defined function. be sure to specify the type of endpoint each piece of the function will have and why.
f(x)=\begin{cases}-x + 3, & x<2\\3, & 2leq x<4\\4 - 2x, & xgeq4end{cases}
graph the three pieces as described above, with the appropriate open and closed endpoints for each piece based on the domain inequalities.
sample response: you would graph the equation (f(x)=-x + 3) for input values less than 2. there would be an open circle at the point ((2,1)) since the domain for the first piece does not include 2. you would then graph a horizontal line at (f(x)=3) for input values between 2 and 4. there would be a closed circle at ((2,3)) and an open circle at ((4,3)). last, you would graph (f(x)=4 - 2x) for input values greater than or equal to 4. there would be a closed circle at the point ((4,-4)) since 4 is in the domain of the third piece.
what did you include in your response? check all that apply.
□ there would be an open circle at ((2,1)).
□ there would be a closed circle at ((2,3)).
□ there would be an open circle at ((4,3)).
□ there would be a closed circle at ((4,-4)).

Explanation:

For the piece - wise function, when \(x < 2\), \(f(x)=-x + 3\), and since \(x = 2\) is not in this domain, there is an open - circle at \((2,1)\) (by substituting \(x = 2\) into \(-x + 3\)). For \(2\leq x<4\), \(f(x)=3\), so there is a closed - circle at \((2,3)\) (because \(x = 2\) is included) and an open - circle at \((4,3)\) (because \(x = 4\) is not included). For \(x\geq4\), \(f(x)=4 - 2x\), and there is a closed - circle at \((4,-4)\) (since \(x = 4\) is in the domain).

Answer:

There would be an open circle at \((2,1)\).
There would be a closed circle at \((2,3)\).
There would be an open circle at \((4,3)\).
There would be a closed circle at \((4,-4)\).