We are given the following functions and their composition:

• $f(x) = ax + b$
• $g(x) = 1 - x + x^2$
• $g(f(x)) = 9x^2 + 9x + 3$

The goal is to determine $a$ and $b$.

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

First, substitute $f(x) = ax + b$ into $g(x)$. Since $g(x) = 1 - x + x^2$, we get:

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

Now expand each term:

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

For the square term $(ax + b)^2$, use the formula $(p + q)^2 = p^2 + 2pq + q^2$:

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

Substituting this back into the equation for $g(f(x))$, we have:

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

Simplify the expression:

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

Step 2: Set $g(f(x))$ equal to $9x^2 + 9x + 3$

We are told that $g(f(x)) = 9x^2 + 9x + 3$. So, equating the expressions for $g(f(x))$:

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

Step 3: Compare coefficients

Now, compare the coefficients of $x^2$, $x$, and the constant terms on both sides:

1. Coefficient of $x^2$:
   $a^2 = 9$
   Solving this gives $a = 3$ or $a = -3$.

2. Coefficient of $x$:
   $2ab - a = 9$

3. Constant term:
   $1 - b + b^2 = 3$

Step 4: Solve for $b$

Let’s first solve for $b$ by examining the constant term equation:

$1 - b + b^2 = 3$ $b^2 - b - 2 = 0$

This is a quadratic equation, and we can solve it using the quadratic formula:

$b = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-2)}}{2(1)}$ $b = \frac{1 \pm \sqrt{1 + 8}}{2}$ $b = \frac{1 \pm 3}{2}$

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

Step 5: Solve for $a$

Now, let’s substitute these values of $b$ into the equation for the coefficient of $x$:

$2ab - a = 9$

Case 1: $a = 3$

• If $b = 2$: $2(3)(2) - 3 = 9 \quad \text{(True)}$ So, $a = 3$ and $b = 2$ is a valid solution.

• If $b = -1$: $2(3)(-1) - 3 = -9 \quad \text{(False)}$ So, $b = -1$ does not work with $a = 3$.

Case 2: $a = -3$