Question

f(x) = \

$$\begin{cases} x^2 - 1, & x < 0 \\\\ 2x, & 0 \\leq x \\leq 4 \\\\ x + 3, & x > 4 \\end{cases}$$

problems (evaluate)

1. f(0) =
2. f(1) =
3. f(4) =
4. f(6) =

bonus (optional)
graph one of the piecewise functions above using desmos. clearly label all open and closed circles.

Explanation:

To solve the piecewise function \( f(x) =

$$\begin{cases} x^2 - 1, & x < 0 \\ 2x, & 0 \leq x \leq 4 \\ x + 3, & x > 4 \end{cases}$$

\) for different values of \( x \), we determine which piece of the function to use based on the value of \( x \).

Problem 7: Evaluate \( f(0) \)

Step 1: Determine the correct piece

Since \( x = 0 \) and the condition \( 0 \leq x \leq 4 \) is satisfied, we use the piece \( 2x \).

Step 2: Substitute \( x = 0 \) into \( 2x \)

\( f(0) = 2(0) = 0 \)

Problem 8: Evaluate \( f(1) \)

Step 1: Determine the correct piece

Since \( x = 1 \) and the condition \( 0 \leq x \leq 4 \) is satisfied, we use the piece \( 2x \).

Step 2: Substitute \( x = 1 \) into \( 2x \)

\( f(1) = 2(1) = 2 \)

Problem 9: Evaluate \( f(4) \)

Step 1: Determine the correct piece

Since \( x = 4 \) and the condition \( 0 \leq x \leq 4 \) is satisfied, we use the piece \( 2x \).

Step 2: Substitute \( x = 4 \) into \( 2x \)

\( f(4) = 2(4) = 8 \)

Problem 10: Evaluate \( f(6) \)

Step 1: Determine the correct piece

Since \( x = 6 \) and the condition \( x > 4 \) is satisfied, we use the piece \( x + 3 \).

Step 2: Substitute \( x = 6 \) into \( x + 3 \)

\( f(6) = 6 + 3 = 9 \)

Answer:

s:

1. \( f(0) = \boldsymbol{0} \)
2. \( f(1) = \boldsymbol{2} \)
3. \( f(4) = \boldsymbol{8} \)
4. \( f(6) = \boldsymbol{9} \)