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 $-12$.

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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.