We are tasked with finding the exact point on the curve $y = 9 + 2e^{-x} - 3x$ where the tangent line is parallel to the line $3x - y = 2$.

Step 1: Find the slope of the given line.

To determine the slope of the line $3x - y = 2$, rewrite it in slope-intercept form, $y = mx + b$, where $m$ is the slope:

$3x - y = 2 \implies y = 3x - 2$

Thus, the slope of the given line is $m = 3$.

Step 2: Find the derivative of the curve.

To find where the tangent line is parallel, we need the derivative of the curve $y = 9 + 2e^{-x} - 3x$, which will give us the slope of the tangent line at any point:

$y' = \frac{d}{dx} \left( 9 + 2e^{-x} - 3x \right)$ $y' = 0 - 2e^{-x} - 3 = -2e^{-x} - 3$

Step 3: Set the derivative equal to the slope of the line.

Since the tangent line is parallel to the given line, we set the derivative equal to the slope of the line, $m = 3$:

$-2e^{-x} - 3 = 3$

Solving for $x$:

$-2e^{-x} = 6$ $e^{-x} = -3 \quad (\text{no real solution since } e^{-x} \text{ is always positive}).$

This means there is no real solution to this equation, and no real point exists on the given curve where the tangent line is parallel to the line $3x - y = 2$.

Would you like more details or have any further questions? Here are some follow-up queries:

1. How is the derivative related to the slope of a tangent line?
2. Can you explain how we rewrote the line equation to find its slope?
3. What does it mean if an equation has no real solution in this context?
4. How do exponential functions behave when exponent is negative?
5. What other methods could we use to confirm that no real solution exists?

Tip: For equations involving exponential functions, be mindful of their range—exponential functions are always positive for real values of the exponent.