find the values of a and b that make f continuous everywhere.\nf(x)=\begin{cases}\frac{x^{2}-4}{x…

Explanation:

Step1: Simplify the first - piece function

For (x < 2), (f(x)=\frac{x^{2}-4}{x - 2}=\frac{(x + 2)(x - 2)}{x - 2}=x + 2) (for (x\neq2)). The limit as (x\to2^{-}) is (\lim_{x\to2^{-}}f(x)=\lim_{x\to2^{-}}(x + 2)=4).

Step2: Calculate the left - hand and right - hand limits at (x = 2)

For (2\leq x<3), (f(x)=ax^{2}-bx + 3). The limit as (x\to2^{+}) is (\lim_{x\to2^{+}}f(x)=4a-2b + 3). For the function to be continuous at (x = 2), (4a-2b+3 = 4), which simplifies to (4a-2b=1).

Step3: Calculate the left - hand and right - hand limits at (x = 3)

For (2\leq x<3), (\lim_{x\to3^{-}}f(x)=9a-3b + 3). For (x\geq3), (f(x)=4x-a + b), and (\lim_{x\to3^{+}}f(x)=12-a + b). For the function to be continuous at (x = 3), (9a-3b+3=12-a + b), which simplifies to (10a-4b=9).

Step4: Solve the system of equations

We have the system of equations (\begin{cases}4a-2b=1\10a-4b=9\end{cases}). Multiply the first equation by (2) to get (8a-4b = 2). Subtract this from the second equation: ((10a-4b)-(8a - 4b)=9 - 2), (2a=7), so (a=\frac{7}{2}). Substitute (a=\frac{7}{2}) into (4a-2b=1): (4\times\frac{7}{2}-2b=1), (14-2b=1), (2b = 13), (b=\frac{13}{2}).

Answer:

(a=\frac{7}{2}), (b=\frac{13}{2})