For the following​ function, make a table of slopes of secant lines and make a conjecture about the slope of the tangent line at the indicated point.
​f(x)equals=12 cosine x12cosx at xequals=StartFraction pi Over 2 EndFraction
π
2
Question content area bottom
Part 1
Complete the table below.
​(Round the final answer to three decimal places as needed. Round all intermediate values to four decimal places as​ needed.)
Part 2Part 3Part 4
Interval
Slope of secant line
left bracket StartFraction pi Over 2 EndFraction comma pi right bracket
π
2,π
negative 7.639−7.639
left bracket StartFraction pi Over 2 EndFraction comma StartFraction pi Over 2 EndFraction plus 0.1 right bracket
π
2,
π
2+0.1
negative 11.98−11.98
left bracket StartFraction pi Over 2 EndFraction comma StartFraction pi Over 2 EndFraction plus 0.01 right bracket
π
2,
π
2+0.01
negative 12−12
left bracket StartFraction pi Over 2 EndFraction comma StartFraction pi Over 2 EndFraction plus 0.001 right bracket
π
2,
π
2+0.001
enter your response here


We are given the function \( f(x) = 12 \cos(x) \), and we're tasked with calculating the slopes of secant lines over increasingly smaller intervals near \( x = \frac{\pi}{2} \) to conjecture the slope of the tangent line at that point. 

### Step-by-step outline:

1. The slope of the secant line between two points \( x_1 \) and \( x_2 \) is given by the difference quotient:
   \[
   \text{slope of secant line} = \frac{f(x_2) - f(x_1)}{x_2 - x_1}
   \]
   In our case, \( f(x_1) = f\left( \frac{\pi}{2} \right) = 12 \cos\left( \frac{\pi}{2} \right) = 12(0) = 0 \).
   So, we are calculating:
   \[
   \text{slope} = \frac{f(x_2) - 0}{x_2 - \frac{\pi}{2}} = \frac{f(x_2)}{x_2 - \frac{\pi}{2}}
   \]
   for different values of \( x_2 \).

### Part 1: Compute the slopes for the given intervals:

1. **Interval \( \left[ \frac{\pi}{2}, \pi \right] \):**
   \[
   x_2 = \pi, \quad f(\pi) = 12 \cos(\pi) = 12(-1) = -12
   \]
   \[
   \text{slope} = \frac{-12}{\pi - \frac{\pi}{2}} = \frac{-12}{\frac{\pi}{2}} = -\frac{12 \times 2}{\pi} \approx -7.639
   \]

2. **Interval \( \left[ \frac{\pi}{2}, \frac{\pi}{2} + 0.1 \right] \):**
   \[
   x_2 = \frac{\pi}{2} + 0.1, \quad f\left( \frac{\pi}{2} + 0.1 \right) = 12 \cos\left( \frac{\pi}{2} + 0.1 \right) \approx 12 \cos(1.6708) \approx 12(-0.0998) \approx -1.1976
   \]
   \[
   \text{slope} = \frac{-1.1976}{\left( \frac{\pi}{2} + 0.1 \right) - \frac{\pi}{2}} = \frac{-1.1976}{0.1} = -11.976
   \]
   Rounded to three decimal places: \( -11.980 \).

3. **Interval \( \left[ \frac{\pi}{2}, \frac{\pi}{2} + 0.01 \right] \):**
   \[
   x_2 = \frac{\pi}{2} + 0.01, \quad f\left( \frac{\pi}{2} + 0.01 \right) = 12 \cos\left( \frac{\pi}{2} + 0.01 \right) \approx 12 \cos(1.5808) \approx 12(-0.0099998) \approx -0.12
   \]
   \[
   \text{slope} = \frac{-0.12}{0.01} = -12
   \]

4. **Interval \( \left[ \frac{\pi}{2}, \frac{\pi}{2} + 0.001 \right] \):**
   \[
   x_2 = \frac{\pi}{2} + 0.001, \quad f\left( \frac{\pi}{2} + 0.001 \right) = 12 \cos\left( \frac{\pi}{2} + 0.001 \right) \approx 12 \cos(1.5718) \approx 12(-0.001) \approx -0.012
   \]
   \[
   \text{slope} = \frac{-0.012}{0.001} = -12
   \]

### Part 2: Table of slopes of secant lines

| Interval                                             | Slope of Secant Line |
|------------------------------------------------------|----------------------|
| \( \left[ \frac{\pi}{2}, \pi \right] \)              | \( -7.639 \)         |
| \( \left[ \frac{\pi}{2}, \frac{\pi}{2} + 0.1 \right] \) | \( -11.980 \)        |
| \( \left[ \frac{\pi}{2}, \frac{\pi}{2} + 0.01 \right] \) | \( -12 \)            |
| \( \left[ \frac{\pi}{2}, \frac{\pi}{2} + 0.001 \right] \) | \( -12 \)            |

### Part 3: Conjecture
As the interval gets smaller, the slopes of the secant lines approach \( -12 \). Therefore, the slope of the tangent line at \( x = \frac{\pi}{2} \) is likely \( -12 \).

---

Would you like further details or have any questions?

Here are five related questions for further exploration:

1. What is the significance of the derivative at a point in terms of the tangent line?
2. How does the difference quotient relate to the concept of a derivative?
3. Can we compute the exact derivative of \( f(x) = 12 \cos(x) \) using the definition of a derivative?
4. How do secant line slopes differ from tangent line slopes geometrically?
5. What happens to the slope of the secant line if we move the point of interest to \( x = 0 \)?

**Tip:** The slope of a tangent line at a point is equivalent to the derivative of the function at that point.

For the following​ function, make a table of slopes of secant lines and make a conjecture about the slope of the tangent line at the indicated point. ​f(x) = 12 cos(x) at x = π/2.

Explore how to calculate the slopes of secant lines for the function f(x) = 12 cos(x) at x = π/2. This calculus problem demonstrates the difference quotient method to approximate the slope of the tangent line. Suitable for advanced high school or college-level calculus students.

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