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Let's break down the problem step-by-step to find \( p + q \).

Given:
- \( f(x) = \frac{1}{2}x^2 \)
- \( g(x) = x^3 \)

1. **Find the tangent line to \( f(x) \):**
   - The derivative of \( f(x) \) is \( f'(x) = x \).
   - The tangent line at point \( (a, b) \) on \( f(x) \) has the equation:
     \[
     y - f(a) = f'(a)(x - a)
     \]
     Substitute \( f(a) = \frac{1}{2}a^2 \) and \( f'(a) = a \):
     \[
     y - \frac{1}{2}a^2 = a(x - a)
     \]
     Simplify this to get the equation of the tangent line \( L \):
     \[
     y = ax - \frac{1}{2}a^2
     \]

2. **Find the point where this line is also tangent to \( g(x) \):**
   - The derivative of \( g(x) \) is \( g'(x) = 3x^2 \).
   - For \( L \) to be tangent to \( g(x) \) at some point \( (c, d) \):
     1. The line must pass through \( (c, g(c)) \):
        \[
        g(c) = c^3
        \]
        \[
        c^3 = ac - \frac{1}{2}a^2
        \]
     2. The slopes must match:
        \[
        g'(c) = 3c^2 = a
        \]

3. **Express \( a \) in terms of \( c \):**
   \[
   a = 3c^2
   \]

4. **Substitute into the point condition:**
   \[
   c^3 = (3c^2)c - \frac{1}{2}(3c^2)^2
   \]
   Simplify the right side:
   \[
   c^3 = 3c^3 - \frac{1}{2} \times 9c^4
   \]
   \[
   c^3 = 3c^3 - \frac{9}{2}c^4
   \]
   Combine like terms:
   \[
   0 = 2c^3 - \frac{9}{2}c^4
   \]
   Factor out \( c^3 \):
   \[
   c^3 \left(2 - \frac{9}{2}c \right) = 0
   \]
   This gives:
   - \( c^3 = 0 \) which implies \( c = 0 \) (but this is not valid since \( c \neq 0 \) by the problem's condition)
   - \( 2 - \frac{9}{2}c = 0 \)

5. **Solve for \( c \):**
   \[
   2 = \frac{9}{2}c
   \]
   \[
   c = \frac{4}{9}
   \]

6. **Find \( a \) using \( a = 3c^2 \):**
   \[
   a = 3 \left(\frac{4}{9}\right)^2 = 3 \times \frac{16}{81} = \frac{48}{81} = \frac{16}{27}
   \]

7. **Find \( b \) and \( d \):**
   - \( b = f(a) = \frac{1}{2}a^2 = \frac{1}{2} \left(\frac{16}{27}\right)^2 = \frac{1}{2} \times \frac{256}{729} = \frac{128}{729} \)
   - \( d = g(c) = \left(\frac{4}{9}\right)^3 = \frac{64}{729} \)

8. **Calculate \( b + d + \frac{ac}{3} \):**
   \[
   b + d = \frac{128}{729} + \frac{64}{729} = \frac{192}{729}
   \]
   \[
   \frac{ac}{3} = \frac{\left(\frac{16}{27}\right) \left(\frac{4}{9}\right)}{3} = \frac{\frac{64}{243}}{3} = \frac{64}{729}
   \]
   Combine everything:
   \[
   b + d + \frac{ac}{3} = \frac{192}{729} + \frac{64}{729} = \frac{256}{729}
   \]

9. **Express as \( \frac{p}{q} \) and compute \( p + q \):**
   \[
   \frac{256}{729} \quad \Rightarrow \quad p = 256, \quad q = 729
   \]
   \[
   p + q = 256 + 729 = 985
   \]

Thus, the final answer is \( p + q = 985 \).

Would you like a detailed breakdown of any step, or do you have any questions? 

Here are 5 related questions to deepen understanding:
1. How do you find the equation of a tangent line for a given function?
2. Why must the slopes of the tangent lines be equal for the line to be tangent to both functions?
3. How would this problem change if the functions \( f(x) \) and \( g(x) \) were different?
4. How can the condition \( c \neq 0 \) change the solution set of this problem?
5. What is the geometric interpretation of a line being tangent to two different curves?

**Tip:** Always double-check the conditions given in a problem, such as \( c \neq 0 \) here, to ensure solutions are valid.

Let f(x) = (1/2)x^2 and g(x) = x^3. Let L be a tangent line that contacts a curve y = f(x) at (a, b). L also contacts a curve of y = g(x) at (c, d). Let b + d + ac/3 = p/q where a, b, c, d ∈ R and c ≠ 0. Compute p + q.

Learn how to compute the intersection of tangent lines for the curves f(x) = (1/2)x^2 and g(x) = x^3. This problem involves derivatives, tangent lines, and solving for values such as p and q to find their sum.

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