17. [-/2 Points] DETAILS \( \begin{array}{l}\text { MY NOTES } \\ \text { Find the values of } a \text { and } b \text { that make } f \text { continuous everywhere. } \\ \qquad f(x)=\left\{\begin{array}{ll}\frac{x^{2}-4}{x-2} & \text { if } x<2 \\ a x^{2}-b x+3 & \text { if } 2 \leq x<3 \\ 2 x-a+b & \text { if } x \geq 3\end{array}\right. \\ a=\square \\ b=\square\end{array} \)

To ensure that the function \( f(x) \) is continuous everywhere, we need to ensure that both the left-hand limit and the right-hand limit at the points of transition (specifically at \( x=2 \) and \( x=3 \)) match the function's value. 1. For \( x < 2 \), we can simplify \( f(x) \) as: \[ f(x) = \frac{x^2 - 4}{x - 2} = x + 2 \quad \text{(as long as \( x \neq 2 \))} \] Evaluating this as \( x \) approaches 2 gives: \[ \lim_{x \to 2^-} f(x) = 2 + 2 = 4 \] Moving to \( x = 2 \) for the second piece, we set \( f(2) \) equal to the left limit: \[ a(2)^2 - b(2) + 3 = 4 \] Simplifying, we get: \[ 4a - 2b + 3 = 4 \\ \Rightarrow 4a - 2b = 1 \quad \text{(Equation 1)} \] 2. Next, we consider the transition at \( x = 3 \): For the second piece, evaluating at 3 gives: \[ a(3)^2 - b(3) + 3 = 9a - 3b + 3 \] The right-hand limit from the third piece \( f(x) = 2x - a + b \) at \( x=3 \) is: \[ 2(3) - a + b = 6 - a + b \] Setting these equal for continuity: \[ 9a - 3b + 3 = 6 - a + b \\ \Rightarrow 10a - 4b + 3 = 6 \\ \Rightarrow 10a - 4b = 3 \quad \text{(Equation 2)} \] Now we solve these two simultaneous equations: From Equation 1: \[ 2b = 4a - 1 \\ b = 2a - 0.5 \] Substituting for \( b \) in Equation 2: \[ 10a - 4(2a - 0.5) = 3 \\ 10a - 8a + 2 = 3 \\ 2a = 1 \\ a = \frac{1}{2} \] Putting \( a \) back into the expression for \( b \): \[ b = 2\left(\frac{1}{2}\right) - 0.5 = 1 - 0.5 = 0.5 \] So, the final values are: \[ a = \frac{1}{2}, \quad b = \frac{1}{2} \]