Question

piecewise function (problems 7-10)
$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(5) = $ ______

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

Explanation:

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

Step1: Determine the domain for \( x = 0 \)

Given the piecewise function, for \( x = 0 \), we check the intervals. The interval \( 0 \leq x \leq 4 \) includes \( x = 0 \), so we use the function \( f(x)=2x \) for this domain.

Step2: Substitute \( x = 0 \) into \( f(x)=2x \)

Substitute \( x = 0 \) into \( f(x)=2x \), we get \( f(0)=2\times0 = 0 \).

Step1: Determine the domain for \( x = 1 \)

For \( x = 1 \), it lies in the interval \( 0 \leq x \leq 4 \), so we use the function \( f(x)=2x \).

Step2: Substitute \( x = 1 \) into \( f(x)=2x \)

Substitute \( x = 1 \) into \( f(x)=2x \), we get \( f(1)=2\times1 = 2 \).

Step1: Determine the domain for \( x = 4 \)

For \( x = 4 \), the interval \( 0 \leq x \leq 4 \) includes \( x = 4 \) (since it's a closed interval at 4), so we use the function \( f(x)=2x \).

Step2: Substitute \( x = 4 \) into \( f(x)=2x \)

Substitute \( x = 4 \) into \( f(x)=2x \), we get \( f(4)=2\times4 = 8 \).

Answer:

\( f(0)=0 \)

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