Question

f(x)=\

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

evaluate f(0)

Explanation:

Step1: Determine the applicable function part

Since we need to evaluate \( f(0) \), we check the conditions. The first part of the piece - wise function is \( 3x + 1 \) with the condition \( x\leq0 \). Since \( 0\leq0 \), we use the function \( f(x)=3x + 1 \) for \( x = 0 \).

Step2: Substitute \( x = 0 \) into the function

Substitute \( x = 0 \) into \( f(x)=3x + 1 \). We get \( f(0)=3(0)+1 \).

Step3: Calculate the result

First, calculate \( 3(0)=0 \). Then, \( 0 + 1=1 \).

Answer:

\( 1 \)