Question

find the values of a and b that make f continuous everywhere.

f(x) =
\

$$\begin{cases} \\dfrac{x^2 - 4}{x - 2} & \\text{if } x < 2 \\\\ ax^2 - bx + 1 & \\text{if } 2 \\leq x < 3 \\\\ 4x - a + b & \\text{if } x \\geq 3 \\end{cases}$$

a = \boxed{}

b = \boxed{}

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Explanation:

Step1: Simplify the first piece at \(x = 2\)

For \(x<2\), \(f(x)=\frac{x^{2}-4}{x - 2}\). We can factor the numerator: \(x^{2}-4=(x - 2)(x + 2)\). So \(\frac{(x - 2)(x + 2)}{x - 2}=x + 2\) (for \(x
eq2\)). The limit as \(x\to2^{-}\) is \(\lim_{x\to2^{-}}(x + 2)=2+2 = 4\).

For the function to be continuous at \(x = 2\), the limit as \(x\to2^{+}\) (using the second piece \(ax^{2}-bx + 1\)) must equal 4. So when \(x = 2\), \(a(2)^{2}-b(2)+1=4\), which simplifies to \(4a-2b+1 = 4\), or \(4a-2b=3\) (Equation 1).

Step2: Find the limit at \(x = 3\) for both sides

For the second piece, when \(x\to3^{-}\), \(f(x)=ax^{2}-bx + 1\), so \(\lim_{x\to3^{-}}f(x)=a(3)^{2}-b(3)+1=9a-3b + 1\).

For the third piece, when \(x\to3^{+}\), \(f(x)=4x-a + b\), so \(\lim_{x\to3^{+}}f(x)=4(3)-a + b=12-a + b\).

For continuity at \(x = 3\), \(9a-3b + 1=12-a + b\). Simplify this: \(9a+a-3b - b=12 - 1\), so \(10a-4b=11\) (Equation 2).

Step3: Solve the system of equations

We have Equation 1: \(4a-2b=3\) (we can multiply this by 2 to get \(8a - 4b=6\)) and Equation 2: \(10a-4b=11\).

Subtract the new Equation 1 from Equation 2: \((10a-4b)-(8a - 4b)=11 - 6\).

Simplify: \(10a-4b - 8a + 4b=5\), so \(2a=5\), which gives \(a=\frac{5}{2}=2.5\).

Now substitute \(a=\frac{5}{2}\) into Equation 1: \(4(\frac{5}{2})-2b=3\).

Simplify: \(10-2b=3\), then \(-2b=3 - 10=-7\), so \(b=\frac{7}{2}=3.5\).

Answer:

\(a=\frac{5}{2}\), \(b=\frac{7}{2}\) (or \(a = 2.5\), \(b=3.5\))