Find the values of \( a \) and \( b \) that make \( f \) continuous everywhere. \[ f(x)=\left\{\begin{array}{ll}\frac{x^{2}-4}{x-2} & \text { if } x<2 \\ a x^{2}-b x+3 & \text { if } 2 \leq x<3 \\ 4 x-a+b & \text { if } x \geq 3\end{array}\right. \] \( a= \) \( b= \)

To find the values of \( a \) and \( b \) that make the function \( f \) continuous everywhere, we need to check the continuity at the points where the definition of \( f(x) \) changes, specifically at \( x = 2 \) and \( x = 3 \). **Step 1: Check continuity at \( x = 2 \)** For \( x < 2 \): \[ f(x) = \frac{x^{2}-4}{x-2} \] This can be simplified: \[ f(x) = \frac{(x-2)(x+2)}{x-2} = x + 2 \quad \text{for } x < 2 \text{ (valid for } x \neq 2\text{)}. \] Thus, as \( x \) approaches 2 from the left: \[ \lim_{x \to 2^-} f(x) = 2 + 2 = 4. \] For \( 2 \leq x < 3 \): \[ f(2) = a(2^2) - b(2) + 3 = 4a - 2b + 3. \] We want the left-hand limit to equal the value of the function at \( x = 2 \): \[ 4a - 2b + 3 = 4. \] Simplifying this, we get: \[ 4a - 2b = 1. \tag{1} \] **Step 2: Check continuity at \( x = 3 \)** For \( 2 \leq x < 3 \): \[ f(3) = a(3^2) - b(3) + 3 = 9a - 3b + 3. \] For \( x \geq 3 \): \[ f(3) = 4(3) - a + b = 12 - a + b. \] We want the left-hand value at \( x = 3 \) to equal the right-hand one: \[ 9a - 3b + 3 = 12 - a + b. \] Rearranging gives us: \[ 9a - 3b + 3 + a - b = 12. \] Combining like terms: \[ 10a - 4b = 9. \tag{2} \] **Step 3: Solve the system of equations** We have two equations: 1. \( 4a - 2b = 1 \) 2. \( 10a - 4b = 9 \) From equation (1), we can express \( b \) in terms of \( a \): \[ 2b = 4a - 1 \implies b = 2a - \frac{1}{2}. \] Substituting \( b \) into equation (2): \[ 10a - 4(2a - \frac{1}{2}) = 9. \] This simplifies to: \[ 10a - 8a + 2 = 9 \implies 2a + 2 = 9 \implies 2a = 7 \implies a = \frac{7}{2}. \] Now substitute \( a \) back into the expression for \( b \): \[ b = 2\left(\frac{7}{2}\right) - \frac{1}{2} = 7 - \frac{1}{2} = 6.5 = \frac{13}{2}. \] Thus, the values of \( a \) and \( b \) that make \( f \) continuous everywhere are: \[ a = \frac{7}{2}, \quad b = \frac{13}{2}. \] Final Answer: \[ a = \frac{7}{2}, \qquad b = \frac{13}{2}. \]