Answer :
Sure! Let's evaluate the piecewise function at the given points step-by-step.
The piecewise function in question is defined as:
[tex]\[ f(x) = \begin{cases} 5 & \text{if } x \leq -1 \\ 2x + 1 & \text{if } 1 < x < 3 \\ 7 & \text{if } x \geq 3 \end{cases} \][/tex]
Step 1: Evaluating [tex]\( f(-3) \)[/tex]
Since [tex]\( -3 \leq -1 \)[/tex], we use the first part of the piecewise function:
[tex]\[ f(-3) = 5 \][/tex]
Step 2: Evaluating [tex]\( f(-1) \)[/tex]
Since [tex]\( -1 \leq -1 \)[/tex], it falls under the same case as the previous step:
[tex]\[ f(-1) = 5 \][/tex]
Step 3: Evaluating [tex]\( f(3) \)[/tex]
Since [tex]\( 3 \geq 3 \)[/tex], we use the last part of the piecewise function:
[tex]\[ f(3) = 7 \][/tex]
Summarizing the evaluations:
[tex]\[ \begin{align*} f(-3) &= 5, \\ f(-1) &= 5, \\ f(3) &= 7. \end{align*} \][/tex]
Thus, the values are:
[tex]\[ \begin{array}{l} f(-3) = 5 \\ f(-1) = 5 \\ f(3) = 7 \end{array} \][/tex]
The piecewise function in question is defined as:
[tex]\[ f(x) = \begin{cases} 5 & \text{if } x \leq -1 \\ 2x + 1 & \text{if } 1 < x < 3 \\ 7 & \text{if } x \geq 3 \end{cases} \][/tex]
Step 1: Evaluating [tex]\( f(-3) \)[/tex]
Since [tex]\( -3 \leq -1 \)[/tex], we use the first part of the piecewise function:
[tex]\[ f(-3) = 5 \][/tex]
Step 2: Evaluating [tex]\( f(-1) \)[/tex]
Since [tex]\( -1 \leq -1 \)[/tex], it falls under the same case as the previous step:
[tex]\[ f(-1) = 5 \][/tex]
Step 3: Evaluating [tex]\( f(3) \)[/tex]
Since [tex]\( 3 \geq 3 \)[/tex], we use the last part of the piecewise function:
[tex]\[ f(3) = 7 \][/tex]
Summarizing the evaluations:
[tex]\[ \begin{align*} f(-3) &= 5, \\ f(-1) &= 5, \\ f(3) &= 7. \end{align*} \][/tex]
Thus, the values are:
[tex]\[ \begin{array}{l} f(-3) = 5 \\ f(-1) = 5 \\ f(3) = 7 \end{array} \][/tex]
