Given the functions $f(x) = ax + b$ and $g(x) = 1 - x + x^2$, we need to find $a$ and $b$ such that $(g \circ f)(x) = 9x^2 + 9x + 3$.

Step 1: Compute $g(f(x))$

The composition $(g \circ f)(x)$ means we need to substitute $f(x) = ax + b$ into $g(x)$. So, compute $g(f(x))$:

$g(f(x)) = g(ax + b) = 1 - (ax + b) + (ax + b)^2$

Step 2: Expand the expression

Now, expand $(ax + b)^2$:

$(ax + b)^2 = a^2x^2 + 2abx + b^2$

Thus, the full expression for $g(f(x))$ becomes:

$g(f(x)) = 1 - (ax + b) + a^2x^2 + 2abx + b^2$

Simplify the equation:

$g(f(x)) = 1 - ax - b + a^2x^2 + 2abx + b^2$

Group the terms by powers of $x$:

$g(f(x)) = a^2x^2 + (2ab - a)x + (1 - b + b^2)$

Step 3: Compare coefficients

We are given that $g(f(x)) = 9x^2 + 9x + 3$. Now, compare this expression with the expanded form of $g(f(x))$:

$a^2x^2 + (2ab - a)x + (1 - b + b^2) = 9x^2 + 9x + 3$

From this, we can match the coefficients of like terms:

1. Coefficient of $x^2$: $a^2 = 9$
2. Coefficient of $x$: $2ab - a = 9$
3. Constant term: $1 - b + b^2 = 3$

Step 4: Solve the system of equations

1. From $a^2 = 9$, we get $a = 3$ or $a = -3$.

2. Using $a = 3$ or $a = -3$ in the second equation $2ab - a = 9$:

   • If $a = 3$, the equation becomes $6b - 3 = 9$, which simplifies to $6b = 12$, so $b = 2$.
   • If $a = -3$, the equation becomes $-6b + 3 = 9$, which simplifies to $-6b = 6$, so $b = -1$.

3. Substitute the values of $a$ and $b$ into the third equation $1 - b + b^2 = 3$:

   • If $a = 3$ and $b = 2$: $1 - 2 + 2^2 = 1 - 2 + 4 = 3 \quad \text{(True)}$
   • If $a = -3$ and $b = -1$: $1 - (-1) + (-1)^2 = 1 + 1 + 1 = 3 \quad \text{(True)}$

Step 5: Conclusion

Both pairs $(a = 3, b = 2)$ and $(a = -3, b = -1)$ satisfy the equation. Therefore, the possible values for $a$ and $b$ are:

$(a, b) = (3, 2) \quad \text{or} \quad (a, b) = (-3, -1)$

Let me know if you'd like more details or have any questions!

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Here are 5 related questions you might find interesting:

1. How do we solve more complex function compositions?
2. What are other methods to approach solving function compositions?
3. How do function compositions behave with non-polynomial functions?
4. Can you explain how to decompose a composite function into its components?
5. How does the behavior of a quadratic function change with its coefficients?

Tip: When dealing with function compositions, always focus on substituting one function into the other and carefully simplify step-by-step.