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Let's break down and solve each problem shown in the image:

### Question 11:
You are asked to find the equation(s) of the tangent lines to the curve 
\[ y = \frac{x - 1}{x + 1} \]
that are parallel to the line 
\[ x - 2y = 4 \].

#### Step 1: Slope of the given line
We first rewrite the given line equation \( x - 2y = 4 \) in slope-intercept form \( y = mx + b \) to find the slope.
\[
x - 2y = 4 \implies -2y = -x + 4 \implies y = \frac{1}{2}x - 2
\]
So, the slope \( m = \frac{1}{2} \).

#### Step 2: Derivative of the curve
Next, find the derivative \( y' \) of the curve \( y = \frac{x - 1}{x + 1} \) to get the slope of the tangent line.
We use the quotient rule:
\[
y' = \frac{(x+1)(1) - (x-1)(1)}{(x+1)^2} = \frac{(x+1) - (x-1)}{(x+1)^2} = \frac{2}{(x+1)^2}
\]
So, the slope of the tangent line at any point is \( \frac{2}{(x+1)^2} \).

#### Step 3: Set slope equal to \( \frac{1}{2} \)
To find the points where the tangent line has a slope of \( \frac{1}{2} \), we set the derivative equal to \( \frac{1}{2} \):
\[
\frac{2}{(x+1)^2} = \frac{1}{2}
\]
Multiply both sides by \( (x+1)^2 \) and solve for \( x \):
\[
2 = \frac{1}{2}(x+1)^2 \implies (x+1)^2 = 4 \implies x+1 = \pm 2
\]
Thus, \( x = 1 \) or \( x = -3 \).

#### Step 4: Find corresponding \( y \)-values
For \( x = 1 \):
\[
y = \frac{1-1}{1+1} = 0
\]
So, one point is \( (1, 0) \).

For \( x = -3 \):
\[
y = \frac{-3-1}{-3+1} = \frac{-4}{-2} = 2
\]
So, the other point is \( (-3, 2) \).

#### Step 5: Equations of the tangent lines
Now, use the point-slope form \( y - y_1 = m(x - x_1) \) with \( m = \frac{1}{2} \) to find the equations of the tangent lines.

For \( (1, 0) \):
\[
y - 0 = \frac{1}{2}(x - 1) \implies y = \frac{1}{2}x - \frac{1}{2}
\]

For \( (-3, 2) \):
\[
y - 2 = \frac{1}{2}(x + 3) \implies y = \frac{1}{2}x + \frac{7}{2}
\]

So, the equations of the tangent lines are:
\[
y = \frac{1}{2}x - \frac{1}{2}, \quad y = \frac{1}{2}x + \frac{7}{2}
\]

### Question 12:
You are asked to find the derivative and determine the point where the tangent line has a \( y \)-intercept of \( \frac{1}{2} \).

#### Part (a): Find the derivative
The curve is given as:
\[
y = f(x) = \frac{1}{1 + \sqrt{x}}
\]
Use the chain rule to differentiate:
\[
y' = \frac{d}{dx} \left( 1 + \sqrt{x} \right)^{-1}
\]
Let \( u = 1 + \sqrt{x} \), then:
\[
y' = -u^{-2} \cdot \frac{d}{dx} (1 + \sqrt{x}) = -\frac{1}{(1 + \sqrt{x})^2} \cdot \frac{1}{2\sqrt{x}}
\]
Thus, the derivative is:
\[
y' = -\frac{1}{2\sqrt{x}(1 + \sqrt{x})^2}
\]

#### Part (b): Find the point with a given \( y \)-intercept
The tangent line at point \( P = (a, f(a)) \) has a \( y \)-intercept of \( \frac{1}{2} \). We know the equation of a line is \( y = mx + b \), and \( b = \frac{1}{2} \). 

Let me know if you need help continuing from here!

Would you like details on either solution or have any questions? Here are five related questions for practice:
1. How would the solutions change if the slope of the line in Question 11 was different?
2. What is the geometric significance of the derivative in Question 12(a)?
3. How can we determine the concavity of the curve in Question 11?
4. Can the method in Question 11 be generalized to other rational functions?
5. How would we find higher-order derivatives for the function in Question 12(a)?

**Tip:** Always double-check if the points found satisfy the equation of the curve when determining tangent lines!

11. Find equations of the tangent lines to the curve y = (x - 1) / (x + 1) that are parallel to the line x - 2y = 4.
12. Consider the curve y = 1 / (1 + sqrt(x)). (a) Find the derivative f'(x). (b) Find the point on the graph where the tangent line has a y-intercept of 1/2.

This problem involves finding tangent lines to a curve y = (x - 1) / (x + 1) that are parallel to a given line and involves applying the quotient rule for derivatives. Also includes finding the derivative of a function y = 1 / (1 + sqrt(x)) and solving for a point where the tangent has a specific y-intercept.

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