https://file.aiquickdraw.com/mathbot/file/17259182572257fb2p5ou.jpg

Let's break down each of the problems from the image and solve them.

### 1. **Continuity of the function**:
You are given the function \( f(x) = (-0.5 + e^x)^{1/2} \). The question is asking whether this function is continuous at \( x = 0 \).

- For continuity at \( x = 0 \), you need to check if \( \lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0) \).
- The function involves a square root, which may cause discontinuities depending on the expression inside the square root.

   - At \( x = 0 \), \( f(0) = \sqrt{-0.5 + e^0} = \sqrt{-0.5 + 1} = \sqrt{0.5} \), which is defined.
   - As \( x \to 0^+ \) and \( x \to 0^- \), the value inside the square root remains positive, so the function is continuous from both sides at \( x = 0 \).

**Answer**: Continuous from both sides.

---

### 2. **Limit problem**:
You are given \( f(x) = \frac{4x + |x|}{6x - 2|x|} \). You need to determine the limits for \( x \to 0^+ \), \( x \to 0^- \), and \( x \to 0 \).

#### a) \( \lim_{x \to 0^+} f(x) \):
- For \( x \to 0^+ \), \( |x| = x \), so the function simplifies to:
  \[
  f(x) = \frac{4x + x}{6x - 2x} = \frac{5x}{4x} = \frac{5}{4}
  \]
  Therefore, \( \lim_{x \to 0^+} f(x) = \frac{5}{4} \).

#### b) \( \lim_{x \to 0^-} f(x) \):
- For \( x \to 0^- \), \( |x| = -x \), so the function simplifies to:
  \[
  f(x) = \frac{4x - x}{6x + 2x} = \frac{3x}{8x} = \frac{3}{8}
  \]
  Therefore, \( \lim_{x \to 0^-} f(x) = \frac{3}{8} \).

#### c) \( \lim_{x \to 0} f(x) \):
- Since the left-hand limit \( \frac{3}{8} \) and the right-hand limit \( \frac{5}{4} \) are not equal, \( \lim_{x \to 0} f(x) \) does not exist (DNE).

**Answers**:
- \( \lim_{x \to 0^+} f(x) = \frac{5}{4} \)
- \( \lim_{x \to 0^-} f(x) = \frac{3}{8} \)
- \( \lim_{x \to 0} f(x) = \text{DNE} \)

---

### 3. **Rock dropped into a well**:
The rock is dropped into a well 29.4 meters deep. We are asked to find the time it takes to hit the bottom, the velocity upon impact, and the velocity after hitting if the well is lined with sand.

#### a) Time to hit the bottom:
- Using the equation of motion \( h = \frac{1}{2} g t^2 \) where \( h = 29.4 \) m and \( g = 9.8 \) m/s².
  \[
  29.4 = \frac{1}{2} \times 9.8 \times t^2
  \]
  Solving for \( t \):
  \[
  t^2 = \frac{29.4 \times 2}{9.8} = 6 \quad \Rightarrow \quad t = \sqrt{6} \approx 2.45 \text{ seconds}
  \]

#### b) Velocity upon hitting the bottom:
- The velocity is given by \( v = g t \).
  \[
  v = 9.8 \times 2.45 \approx 24.01 \, \text{m/s}
  \]

#### c) Velocity after hitting the bottom (assuming the rock does not bounce):
- If the well is lined with thick sand, the velocity just after hitting the bottom is 0 m/s, because the sand absorbs the impact.

**Answers**:
- \( t \approx 2.45 \, \text{seconds} \)
- \( v \approx 24.01 \, \text{m/s} \)
- \( v_f = 0 \, \text{m/s} \)

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Let me know if you would like further details on any of the steps. 

Here are some questions for further exploration:
1. What conditions could make the function in Problem 1 discontinuous?
2. How does the modulus affect the limits in Problem 2?
3. What is the significance of the left-hand and right-hand limits not matching?
4. Can we find the time it takes for the rock to reach half the depth of the well?
5. What other factors, like air resistance, could change the results of Problem 3?

**Tip**: When solving limit problems involving absolute values, it helps to split the function into different cases for \( x > 0 \) and \( x < 0 \).

The original question submitted by the user as text.

Explore detailed solutions to problems involving continuity of a function, calculating limits, and applying kinematic equations. Suitable for high school students in Grades 11-12.

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