Here are the slopes of the secant lines passing through the point $(1, 2)$ and $(x, f(x))$, rounded to 4 decimal places for each given value of $x$:

• For $x = 0.8$, slope = 2.1115
• For $x = 0.9$, slope = 2.0527
• For $x = 0.99$, slope = 2.0050
• For $x = 0.999$, slope = 2.0005
• For $x = 1.001$, slope = 1.9995
• For $x = 1.01$, slope = 1.9950
• For $x = 1.1$, slope = 1.9524
• For $x = 1.2$, slope = 1.9089

For part (b), based on these slopes, the apparent slope of the tangent line at $x = 1$ appears to be approximately 2.0000.

Would you like more details or have any questions?

Here are five related questions you might find interesting:

1. How do secant lines differ from tangent lines in terms of slope?
2. What happens to the secant line as the second point approaches $(1, 2)$?
3. How can you calculate the derivative of the function $f(x) = 4\sqrt{x} - 2$ at any point?
4. What is the geometric interpretation of the derivative at $x = 1$?
5. How do different values of $x$ affect the accuracy of the slope approximation?

Tip: The closer the points on the secant line are to each other, the closer the secant slope is to the tangent slope.