https://file.aiquickdraw.com/mathbot/file/1729158619953mr76ewro.jpg

Let's analyze the mathematical problem provided in the image.

### Given Function:
\[
f(x) = 
\begin{cases}
4x + 2c & \text{untuk } x < 1 \\
cx^2 + 1 & \text{untuk } x \geq 1
\end{cases}
\]
The question asks to evaluate the limit of \( f(x) \) as \( x \to 1 \) and other conditions related to \( c \).

### Key Concept:
In order for \( f(x) \) to have a limit at \( x = 1 \), both the left-hand limit (as \( x \to 1^- \)) and the right-hand limit (as \( x \to 1^+ \)) must exist and be equal.

1. **Left-hand limit as \( x \to 1^- \)**:
   \[
   \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (4x + 2c) = 4(1) + 2c = 4 + 2c
   \]

2. **Right-hand limit as \( x \to 1^+ \)**:
   \[
   \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (cx^2 + 1) = c(1^2) + 1 = c + 1
   \]

To have the function continuous and a limit at \( x = 1 \), we require:
\[
\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x)
\]
This gives the equation:
\[
4 + 2c = c + 1
\]
Solving for \( c \):
\[
4 + 2c = c + 1 \\
2c - c = 1 - 4 \\
c = -3
\]

### Now let's analyze the statements:

- **a.** \( \lim_{x \to 1^+} f(x) = 10 \) for \( c \) positive: 
  - If \( c \) is positive, the right-hand limit would not be \( 10 \) (as this limit depends on the value of \( c \)).
  - **False.**
  
- **b.** \( \lim_{x \to 1^+} f(x) = 2 \) for \( c \) positive:
  - From \( \lim_{x \to 1^+} f(x) = c + 1 \), if \( c \) is positive, the limit cannot be 2.
  - **False.**

- **c.** \( \lim_{x \to 1^+} f(x) = -5 \) for \( c \) negative:
  - If \( c = -3 \), the right-hand limit is \( \lim_{x \to 1^+} f(x) = -3 + 1 = -2 \), not \( -5 \).
  - **False.**

- **d.** \( \lim_{x \to 1^-} f(x) = 5 \) for \( c \) positive:
  - The left-hand limit depends on \( 4 + 2c \). If \( c = -3 \), the left-hand limit would not be 5. 
  - **False.**

- **e.** \( c^2 + 2c - 3 = 0 \):
  - Solving this quadratic equation:
    \[
    c^2 + 2c - 3 = 0 \implies (c - 1)(c + 3) = 0
    \]
    So, \( c = 1 \) or \( c = -3 \).
  - **True.**

- **f.** \( c^2 - 2c - 3 = 0 \):
  - This quadratic doesn't hold for \( c = -3 \).
  - **False.**

- **g.** \( \lim_{x \to 1^+} f(x) = 2 \) for \( c \) negative:
  - For \( c = -3 \), the right-hand limit is \( -2 \), not 2.
  - **False.**

- **h.** \( \lim_{x \to 1^+} f(x) = -2 \) for \( c \) negative:
  - For \( c = -3 \), the right-hand limit is \( -2 \).
  - **True.**

### Conclusion:
The correct statements are:
- **e.** \( c^2 + 2c - 3 = 0 \)
- **h.** \( \lim_{x \to 1^+} f(x) = -2 \) for \( c \) negative

Would you like further details on the solution or clarification on any point?

---

### Follow-up Questions:
1. What are the conditions for a function to be continuous at a point?
2. Can you explain how limits from the left-hand and right-hand side impact continuity?
3. How do you solve quadratic equations like \( c^2 + 2c - 3 = 0 \)?
4. How would the limits change if \( c = 1 \) instead of \( -3 \)?
5. Why is it necessary for the limits to match at \( x = 1 \) for continuity?

### Tip:
When solving limit problems with piecewise functions, always check both sides of the limit to ensure consistency for continuity.

Evaluate the limits and find the value of c for which the function f(x) is continuous at x = 1. The piecewise function is defined as f(x) = 4x + 2c for x < 1, and f(x) = cx^2 + 1 for x ≥ 1.

Learn how to solve for the value of c that ensures continuity at x = 1 for a piecewise function. This problem covers limits, quadratic equations, and the conditions for continuity. Suitable for students in Grades 11-12.

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