To find the values of $x$ where the curves $y = f(x)$ and $y = g(x)$ have parallel tangent lines, we need to compute the derivatives of $f(x)$ and $g(x)$, and then solve for $x$ such that the derivatives are equal, i.e.,

$f'(x) = g'(x)$

Step 1: Differentiate $f(x)$

Given $f(x) = \frac{x - \frac{3}{2}}{x^2 + 2}$, we can apply the quotient rule:

$f'(x) = \frac{(x^2 + 2) \cdot \left( \frac{d}{dx}(x - \frac{3}{2}) \right) - (x - \frac{3}{2}) \cdot \frac{d}{dx}(x^2 + 2)}{(x^2 + 2)^2}$

Simplifying the derivatives in the numerator: $f'(x) = \frac{(x^2 + 2) \cdot 1 - (x - \frac{3}{2}) \cdot 2x}{(x^2 + 2)^2}$

$f'(x) = \frac{x^2 + 2 - 2x(x - \frac{3}{2})}{(x^2 + 2)^2}$

Expanding and simplifying: $f'(x) = \frac{x^2 + 2 - 2x^2 + 3x}{(x^2 + 2)^2}$ $f'(x) = \frac{-x^2 + 3x + 2}{(x^2 + 2)^2}$

Step 2: Differentiate $g(x)$

Similarly, for $g(x) = \frac{x^2 + 1}{x^2 + 2}$, we apply the quotient rule:

$g'(x) = \frac{(x^2 + 2) \cdot \frac{d}{dx}(x^2 + 1) - (x^2 + 1) \cdot \frac{d}{dx}(x^2 + 2)}{(x^2 + 2)^2}$

Simplifying the derivatives in the numerator: $g'(x) = \frac{(x^2 + 2) \cdot 2x - (x^2 + 1) \cdot 2x}{(x^2 + 2)^2}$ $g'(x) = \frac{2x(x^2 + 2) - 2x(x^2 + 1)}{(x^2 + 2)^2}$

Simplifying further: $g'(x) = \frac{2x(x^2 + 2 - x^2 - 1)}{(x^2 + 2)^2}$ $g'(x) = \frac{2x(1)}{(x^2 + 2)^2} = \frac{2x}{(x^2 + 2)^2}$

Step 3: Solve $f'(x) = g'(x)$

Now, we equate $f'(x)$ and $g'(x)$:

$\frac{-x^2 + 3x + 2}{(x^2 + 2)^2} = \frac{2x}{(x^2 + 2)^2}$

Since both sides have the same denominator, we can set the numerators equal:

$-x^2 + 3x + 2 = 2x$

Simplifying the equation:

$-x^2 + 3x + 2 - 2x = 0$ $-x^2 + x + 2 = 0$

Multiplying through by -1 to simplify:

$x^2 - x - 2 = 0$

Step 4: Solve the quadratic equation

We now solve $x^2 - x - 2 = 0$ using factoring:

$(x - 2)(x + 1) = 0$

So, $x = 2$ or $x = -1$.

Conclusion:

The curves $y = f(x)$ and $y = g(x)$ have parallel tangent lines at $x = 2$ and $x = -1$.

Would you like more details on the steps? Here are 5 related questions you can explore:

1. How do you apply the quotient rule in calculus?
2. What does it mean for two functions to have parallel tangent lines?
3. How can you verify if the solutions $x = 2$ and $x = -1$ are correct graphically?
4. What are other methods to solve quadratic equations?
5. Can two curves have multiple points where their tangent lines are parallel?

Tip: Understanding the geometric interpretation of derivatives can greatly help in visualizing problems involving slopes and tangents.