Let \( f(x)=\left\{\begin{array}{lll}6 x-6 & \text { if } & x \leq 3 \\ -3 x+b & \text { if } & x>3\end{array}\right. \) If \( f(x) \) is a function which is continuous everywhere, then we must have \( b= \)

To ensure that the function \( f(x) \) is continuous everywhere, we need to check the continuity at the point where the definition of the function changes, which is at \( x = 3 \). The function is defined as follows: \[ f(x) = \begin{cases} 6x - 6 & \text{if } x \leq 3 \\ -3x + b & \text{if } x > 3 \end{cases} \] To find the value of \( b \) that makes \( f(x) \) continuous at \( x = 3 \), we need to ensure that the left-hand limit as \( x \) approaches 3 from the left equals the right-hand limit as \( x \) approaches 3 from the right, and both equal \( f(3) \). 1. **Calculate \( f(3) \)**: Since \( 3 \leq 3 \), we use the first case: \[ f(3) = 6(3) - 6 = 18 - 6 = 12 \] 2. **Calculate the left-hand limit as \( x \) approaches 3**: As \( x \) approaches 3 from the left, we again use the first case: \[ \lim_{x \to 3^-} f(x) = 6(3) - 6 = 12 \] 3. **Calculate the right-hand limit as \( x \) approaches 3**: As \( x \) approaches 3 from the right, we use the second case: \[ \lim_{x \to 3^+} f(x) = -3(3) + b = -9 + b \] 4. **Set the limits equal for continuity**: For \( f(x) \) to be continuous at \( x = 3 \), we need: \[ \lim_{x \to 3^-} f(x) = \lim_{x \to 3^+} f(x) \] This gives us the equation: \[ 12 = -9 + b \] 5. **Solve for \( b \)**: Adding 9 to both sides: \[ 12 + 9 = b \implies b = 21 \] Thus, the value of \( b \) that makes \( f(x) \) continuous everywhere is \[ \boxed{21} \]