https://file.aiquickdraw.com/mathbot/file/1726020769852pqvhkbd0.jpg

The piecewise function in the image is written as \( k(x) \) and asks to find the values of \( a \) and \( b \) that make \( k(x) \) continuous everywhere. Here's the breakdown of the function:

\[
k(x) = 
\begin{cases} 
\frac{x^2 - 4}{x - 2} & \text{for } x < 2 \\
ax^2 + bx + 3 & \text{for } 2 \leq x \leq 3 \\
2x - a + b & \text{for } x > 3
\end{cases}
\]

### Continuity Conditions

For \( k(x) \) to be continuous everywhere, it must be continuous at the points where the piecewise function changes, which are at \( x = 2 \) and \( x = 3 \).

#### 1. Continuity at \( x = 2 \)
- The first part is \( \frac{x^2 - 4}{x - 2} \), which simplifies using factorization:

\[
\frac{x^2 - 4}{x - 2} = \frac{(x - 2)(x + 2)}{x - 2} = x + 2 \quad \text{for} \quad x \neq 2
\]

So, as \( x \to 2 \), the limit of the first part is \( 2 + 2 = 4 \).

For continuity at \( x = 2 \), we need \( k(2) \) from the second piece (since the second piece applies at \( x = 2 \)) to equal 4. Therefore, substitute \( x = 2 \) into \( ax^2 + bx + 3 \):

\[
a(2^2) + b(2) + 3 = 4
\]

Simplifying:

\[
4a + 2b + 3 = 4
\]
\[
4a + 2b = 1 \quad \text{(Equation 1)}
\]

#### 2. Continuity at \( x = 3 \)
For continuity at \( x = 3 \), the values of the second and third parts must match at \( x = 3 \).

- Substituting \( x = 3 \) into the second part \( ax^2 + bx + 3 \):

\[
a(3^2) + b(3) + 3 = 9a + 3b + 3
\]

- The third part is \( 2x - a + b \). Substituting \( x = 3 \):

\[
2(3) - a + b = 6 - a + b
\]

For continuity at \( x = 3 \), these two expressions must be equal:

\[
9a + 3b + 3 = 6 - a + b
\]

Simplifying:

\[
9a + 3b + 3 = 6 - a + b
\]
\[
10a + 2b = 3 \quad \text{(Equation 2)}
\]

### Solving the System of Equations
Now, solve the system of equations:

1. \( 4a + 2b = 1 \)
2. \( 10a + 2b = 3 \)

Subtract Equation 1 from Equation 2:

\[
(10a + 2b) - (4a + 2b) = 3 - 1
\]
\[
6a = 2
\]
\[
a = \frac{1}{3}
\]

Substitute \( a = \frac{1}{3} \) into Equation 1:

\[
4\left(\frac{1}{3}\right) + 2b = 1
\]
\[
\frac{4}{3} + 2b = 1
\]
\[
2b = 1 - \frac{4}{3} = \frac{3}{3} - \frac{4}{3} = -\frac{1}{3}
\]
\[
b = -\frac{1}{6}
\]

### Conclusion
The values of \( a \) and \( b \) that make \( k(x) \) continuous everywhere are:
\[
a = \frac{1}{3}, \quad b = -\frac{1}{6}
\]

Would you like a detailed explanation of the next steps or any additional clarification?

Here are 5 related questions for further practice:
1. How do you check if a function is continuous at a point?
2. What is the difference between a piecewise function and a regular function?
3. Can a piecewise function have a discontinuity at a point where the pieces meet?
4. What are the conditions for a function to be differentiable?
5. How would the continuity conditions change if the piecewise function had different intervals?

**Tip:** When solving for continuity in piecewise functions, always check the behavior of the function at the boundaries between different expressions.

Find the values of a and b that make k(x) continuous everywhere for the given piecewise function:

k(x) = 
(x^2 - 4) / (x - 2) for x < 2
ax^2 + bx + 3 for 2 ≤ x ≤ 3
2x - a + b for x > 3


This problem involves solving for the values of a and b in a piecewise function to ensure continuity at x = 2 and x = 3. The function includes quadratic and linear equations, and the solution involves using the definition of continuity and solving systems of equations. Suitable for Grades 11-12.

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