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We are tasked with finding the equation of the tangent line to the function \( f(x) = (3x - x^2)(3 - x - x^2) \) at \( x = 1 \).

### Steps:
1. **Find \( f(1) \)**: This gives the y-coordinate of the point on the graph at \( x = 1 \).
2. **Find \( f'(x) \)**: Differentiate \( f(x) \) using the product rule.
3. **Evaluate \( f'(1) \)**: This gives the slope of the tangent line at \( x = 1 \).
4. **Use the point-slope form of the line equation**: The equation of the tangent line is given by \( y - f(1) = f'(1)(x - 1) \).

Let's calculate the derivative and proceed with the solution.

### Step 1: Find \( f(1) \)
We have:
\[
f(1) = (3(1) - (1)^2)(3 - 1 - (1)^2) = (3 - 1)(3 - 1 - 1) = (2)(1) = 2.
\]

### Step 2: Differentiate \( f(x) \)
We use the product rule for differentiation:
\[
f'(x) = \frac{d}{dx}\left( (3x - x^2)(3 - x - x^2) \right).
\]
Applying the product rule:
\[
f'(x) = (3x - x^2)'(3 - x - x^2) + (3x - x^2)(3 - x - x^2)'.
\]
Now, compute the individual derivatives:
\[
(3x - x^2)' = 3 - 2x,
\]
and
\[
(3 - x - x^2)' = -1 - 2x.
\]

Thus, the derivative becomes:
\[
f'(x) = (3 - 2x)(3 - x - x^2) + (3x - x^2)(-1 - 2x).
\]

### Step 3: Evaluate \( f'(1) \)
Substitute \( x = 1 \) into the expression for \( f'(x) \):
\[
f'(1) = (3 - 2(1))(3 - 1 - 1) + (3(1) - (1)^2)(-1 - 2(1))
= (3 - 2)(1) + (3 - 1)(-1 - 2).
\]
Simplifying:
\[
f'(1) = (1)(1) + (2)(-3) = 1 - 6 = -5.
\]

### Step 4: Equation of the tangent line
The slope of the tangent line at \( x = 1 \) is \( f'(1) = -5 \), and the point on the graph is \( (1, 2) \). Using the point-slope form of the equation of a line:
\[
y - 2 = -5(x - 1).
\]
Simplifying:
\[
y - 2 = -5x + 5,
\]
\[
y = -5x + 7.
\]

Thus, the equation of the tangent line is:
\[
\boxed{y = -5x + 7}.
\]

Would you like more details or have any further questions?

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Here are 5 related questions to expand your understanding:

1. How would the tangent line change if we evaluated it at a different point, say \( x = 0 \)?
2. What does the derivative tell us about the concavity of the function at \( x = 1 \)?
3. Can we find where the tangent line is horizontal? If so, at what points?
4. How can the second derivative help us understand the nature of critical points for \( f(x) \)?
5. Could we apply implicit differentiation if the function was given in a different form?

**Tip**: The point-slope form is useful when you know a point on the line and the slope but don’t want to rearrange into slope-intercept form immediately.

Find the equation of the tangent line to the graph of f(x) = (3x - x^2)(3 - x - x^2) at x = 1.

This problem involves finding the tangent line to the function f(x) = (3x - x^2)(3 - x - x^2) at x = 1 using differentiation and the product rule. The solution covers key calculus concepts such as derivatives, tangent lines, and the product rule.

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