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It appears that you uploaded the same image again, but now you provided an additional detail that the function in the problem is defined as \( kx + 1 \) when \( x \geq 2 \). Let's reconsider the problem with this correction.

### The Function Given:
\[
f(x) = \begin{cases} 
kx + 1 & \text{if } x \geq 2 \\
k^2 + 4k & \text{if } x < 2 
\end{cases}
\]

### Continuity at \( x = 2 \):

To ensure that \( f(x) \) is continuous at \( x = 2 \), the left-hand limit as \( x \) approaches 2 from the left must equal the right-hand limit as \( x \) approaches 2 from the right, and both must equal the value of the function at \( x = 2 \).

### Step 1: Right-hand limit as \( x \) approaches 2 (\( x \geq 2 \))
For \( x \geq 2 \), the function is defined as \( f(x) = kx + 1 \). Thus:
\[
\lim_{x \to 2^+} f(x) = k(2) + 1 = 2k + 1
\]

### Step 2: Left-hand limit as \( x \) approaches 2 (\( x < 2 \))
For \( x < 2 \), the function is defined as \( f(x) = k^2 + 4k \). Thus:
\[
\lim_{x \to 2^-} f(x) = k^2 + 4k
\]

### Step 3: Equating the two limits for continuity
For \( f(x) \) to be continuous at \( x = 2 \), the left-hand limit must equal the right-hand limit:
\[
2k + 1 = k^2 + 4k
\]

### Step 4: Solve the equation
Rearrange the equation to set it to 0:
\[
k^2 + 4k - 2k - 1 = 0 \quad \Rightarrow \quad k^2 + 2k - 1 = 0
\]

This is a quadratic equation, and we can solve for \( k \) using the quadratic formula:
\[
k = \frac{-2 \pm \sqrt{2^2 + 4 \cdot 1 \cdot 1}}{2 \cdot 1} = \frac{-2 \pm \sqrt{4 + 4}}{2} = \frac{-2 \pm \sqrt{8}}{2} = \frac{-2 \pm 2\sqrt{2}}{2} = -1 \pm \sqrt{2}
\]

Thus, the values of \( k \) for which the function \( f(x) \) is continuous are:
\[
k = -1 + \sqrt{2} \quad \text{and} \quad k = -1 - \sqrt{2}
\]

Would you like a detailed explanation of any of these steps?

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Here are some related questions to explore further:

1. What are the specific implications of using \( kx + 1 \) versus \( k - 1 \) in the function definition?
2. How do you determine the continuity of a piecewise function with different polynomial expressions?
3. What happens to the continuity of a function if one of the boundary conditions is changed?
4. Can a function be continuous but not differentiable at a point? Explain with an example.
5. How can solving for \( k \) in piecewise functions be applied to other types of boundary condition problems?

**Tip:** For piecewise functions, always ensure to equate the function values at the boundary points for continuity. This will often involve solving a quadratic equation when parameters are involved.

Explore the continuity of piecewise functions defined by f(x) = kx + 1 for x >= 2. Learn how to determine continuity at boundary points, solve related quadratic equations, and apply these concepts in mathematical problem-solving. Suitable for advanced high school students and those preparing for calculus studies.

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