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To solve the problem and determine the equation of the tangent line to \( f(x) \) at the point \( (0, \frac{1}{2}) \), we will proceed in steps:

### Step 1: Find \( f(0) \)
We need to ensure that the point \( (0, \frac{1}{2}) \) is on the curve by calculating \( f(0) \).

The function is given as:
\[
f(x) = \frac{x^2 + e^x}{2\cos(x) + x}
\]

Substitute \( x = 0 \) into the equation:
\[
f(0) = \frac{0^2 + e^0}{2\cos(0) + 0} = \frac{1}{2 \times 1} = \frac{1}{2}
\]
So, \( f(0) = \frac{1}{2} \), which confirms that the point \( (0, \frac{1}{2}) \) is on the curve.

### Step 2: Find the derivative \( f'(x) \)
Next, we need the slope of the tangent line, which is given by the derivative \( f'(x) \) evaluated at \( x = 0 \).

The function \( f(x) \) is a quotient, so we will use the quotient rule:
\[
f'(x) = \frac{(2x + e^x)(2\cos(x) + x) - (x^2 + e^x)(-2\sin(x) + 1)}{(2\cos(x) + x)^2}
\]

Simplifying each part of this for \( x = 0 \):

- \( 2x + e^x \) becomes \( e^0 = 1 \) at \( x = 0 \).
- \( 2\cos(x) + x \) becomes \( 2 \) at \( x = 0 \).
- \( -2\sin(x) + 1 \) becomes \( 1 \) at \( x = 0 \).

Thus, evaluating the derivative at \( x = 0 \):
\[
f'(0) = \frac{(1 \times 2) - (1 \times 1)}{2^2} = \frac{2 - 1}{4} = \frac{1}{4}
\]
So, the slope of the tangent line at \( x = 0 \) is \( \frac{1}{4} \).

### Step 3: Write the equation of the tangent line
Now that we know the slope is \( \frac{1}{4} \) and the point is \( (0, \frac{1}{2}) \), we use the point-slope form of the equation of a line:
\[
y - y_1 = m(x - x_1)
\]
Substitute \( m = \frac{1}{4} \), \( x_1 = 0 \), and \( y_1 = \frac{1}{2} \):
\[
y - \frac{1}{2} = \frac{1}{4}(x - 0)
\]
Simplify:
\[
y = \frac{1}{4}x + \frac{1}{2}
\]

### Step 4: Compare with answer choices
The correct equation of the tangent line is:
\[
y = \frac{1}{4}x + \frac{1}{2}
\]
This matches option **(D)**.

### Final Answer:
The correct answer is **(D)**.

Would you like more details on any step, or do you have any questions? Here are 5 related questions to deepen your understanding:

1. How does the quotient rule apply to more complex functions like this one?
2. What is the importance of confirming that the given point is on the curve?
3. How can we derive the quotient rule step-by-step from first principles?
4. How do the properties of trigonometric and exponential functions affect the behavior of derivatives?
5. How can you visualize the tangent line in terms of the graph of \( f(x) \)?

**Tip:** Always check if the point given lies on the curve by evaluating the function at the specified \( x \)-value before proceeding with the derivative.

Which of the following is the equation for the tangent line to f(x) = \frac{x^2 + e^x}{2 \cos(x) + x} at the point (0, \frac{1}{2})?

Learn how to find the equation of the tangent line to the function f(x) = (x^2 + e^x) / (2 cos(x) + x) at the point (0, 1/2). This calculus problem involves using the quotient rule to compute the derivative and applying the point-slope form to write the equation of the tangent line.

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