Question

find the values of a and b that make f continuous everywhere.
f(x)=\begin{cases}\frac{x^{2}-4}{x - 2}&\text{if }x<2\ax^{2}-bx + 3&\text{if }2leq x<3\\4x - a + b&\text{if }xgeq3end{cases}
a =
b =

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 \(x
eq2\)). 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:

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