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The problem provided deals with finding the slope of the secant line, estimating the slope of the tangent line, and finally finding the equation of the tangent line to the curve at a given point.

Let's go through each part:

### The curve is given by:
\[
y = \frac{x}{1 + x}
\]

The point \( P(1, \frac{1}{2}) \) lies on this curve.

### (a) Slope of the secant line \( PQ \)

The secant line's slope between the points \( P(1, \frac{1}{2}) \) and \( Q(x, \frac{x}{1+x}) \) is given by:
\[
\text{slope} = \frac{y_Q - y_P}{x_Q - x_P} = \frac{\frac{x}{1 + x} - \frac{1}{2}}{x - 1}
\]

We need to compute this slope for the given values of \( x \):
1. \( x = 0.5 \)
2. \( x = 0.9 \)
3. \( x = 0.99 \)
4. \( x = 0.999 \)
5. \( x = 1.5 \)
6. \( x = 1.1 \)
7. \( x = 1.01 \)
8. \( x = 1.001 \)

I'll compute these slopes numerically to six decimal places. Let me calculate them.Here are the slopes of the secant lines for the given values of \(x\), calculated to six decimal places:

1. \(x = 0.5\): \(0.333333\)
2. \(x = 0.9\): \(0.263158\)
3. \(x = 0.99\): \(0.251256\)
4. \(x = 0.999\): \(0.250125\)
5. \(x = 1.5\): \(0.200000\)
6. \(x = 1.1\): \(0.238095\)
7. \(x = 1.01\): \(0.248756\)
8. \(x = 1.001\): \(0.249875\)

### (b) Estimate of the slope of the tangent line at \(P(1, \frac{1}{2})\):
By observing the slopes as \(x\) approaches 1 from both sides, we can estimate the slope of the tangent line to the curve at \(x = 1\).

It seems that the slopes are approaching \(0.25\) as \(x\) gets closer to 1.

Thus, the estimated slope of the tangent line at \(P(1, \frac{1}{2})\) is:
\[
\boxed{0.25}
\]

### (c) Equation of the tangent line:
Now that we know the slope of the tangent line is \(0.25\), and it passes through the point \(P(1, \frac{1}{2})\), we can use the point-slope form of a line:
\[
y - y_1 = m(x - x_1)
\]
Substitute \(m = 0.25\), \(x_1 = 1\), and \(y_1 = \frac{1}{2}\):
\[
y - \frac{1}{2} = 0.25(x - 1)
\]
Simplifying:
\[
y = 0.25x + 0.25
\]

Thus, the equation of the tangent line is:
\[
\boxed{y = 0.25x + 0.25}
\]

Would you like further clarification on any step or additional details?

### Related Questions:
1. How do you interpret the slope of the tangent line geometrically?
2. What is the derivative of the function \( y = \frac{x}{1 + x} \) at \(x = 1\)?
3. What does the secant line represent in relation to the curve as \(x\) approaches 1?
4. How do you calculate the equation of a tangent line using derivatives?
5. Can you generalize the slope of the tangent line at any point \(x\) on the curve?

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

The point P(1, 1/2) lies on the curve y = x/(1 + x). (a) If Q is the point (x, x/(1 + x)), use your calculator to find the slope of the secant line PQ (correct to six decimal places) for the following values of x: (i) 0.5, (ii) 0.9, (iii) 0.99, (iv) 0.999, (v) 1.5, (vi) 1.1, (vii) 1.01, (viii) 1.001. (b) Using the results of part (a), guess the value of the slope of the tangent line to the curve at P(1, 1/2). (c) Using the slope from part (b), find an equation of the tangent line to the curve at P(1, 1/2).

Learn how to calculate the slope of the secant line and estimate the slope of the tangent line for the curve y = x/(1 + x) at the point P(1, 1/2). This problem demonstrates key calculus concepts such as limits and the tangent line equation. Suitable for students in Grades 10-12.

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