Question

hw 1-8
score: 7.6/15 answered: 8/15
progress saved done
question 9
0/1 pt 3 19 details
part 1 of 2
let ( f(x) = \begin{cases} x^2 - 3 & x < -1 \\ -2x + 4 & -1 leq x < 3 \\sqrt{x - 2} & x geq 3 end{cases} )
create a table for each piece. compare your tables to the options below and make a selection.
a) ( x^2 - 3 ) has select an answer at select an answer
b) ( -2x + 4 ) has select an answer and select an answer
c) ( sqrt{x - 2} ) has select an answer at select an answer

Explanation:

Step1: Analyze $x^2-3$ domain

The piece $x^2 - 3$ applies for $x < -1$. This means it has an open endpoint (no defined value) at $x=-1$. Calculate $f(-1)$ for this piece:
$f(-1) = (-1)^2 - 3 = 1 - 3 = -2$
So this piece has an open circle at $x=-1$.

Step2: Analyze $-2x+4$ domain

The piece $-2x + 4$ applies for $-1 \leq x < 3$. This means it has a closed endpoint at $x=-1$, and an open endpoint at $x=3$. Calculate the values:
$f(-1) = -2(-1) + 4 = 2 + 4 = 6$
$f(3) = -2(3) + 4 = -6 + 4 = -2$
So this piece has a closed circle at $x=-1$ and an open circle at $x=3$.

Step3: Analyze $\sqrt{x-2}$ domain

The piece $\sqrt{x - 2}$ applies for $x \geq 3$. This means it has a closed endpoint at $x=3$. Calculate $f(3)$:
$f(3) = \sqrt{3 - 2} = \sqrt{1} = 1$
So this piece has a closed circle at $x=3$.

Answer:

a) $x^2 - 3$ has an open circle at $x=-1$
b) $-2x + 4$ has a closed circle at $x=-1$ and an open circle at $x=3$
c) $\sqrt{x - 2}$ has a closed circle at $x=3$