Question

f(x)=\

$$\begin{cases}x^3 - 6x, & x \\leq -3 \\\\ -4, & -3 < x \\leq 3 \\\\ \\sqrt{3x - 3}, & x > 3 \\end{cases}$$

evaluate f(4)
type answer here.

Explanation:

Step1: Determine the applicable function

Since \( x = 4 \) and \( 4>3 \), we use the function \( f(x)=\sqrt{3x - 3} \).

Step2: Substitute \( x = 4 \) into the function

Substitute \( x = 4 \) into \( \sqrt{3x - 3} \), we get \( \sqrt{3\times4 - 3} \).

Step3: Simplify the expression

First, calculate the value inside the square root: \( 3\times4 - 3=12 - 3 = 9 \). Then, \( \sqrt{9}=3 \).

Answer:

\( 3 \)