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The problem you provided reads as follows:

**16.**
(a) The function \( f(x) = ax^2 + bx + 3 \), where \( x \geq 2 \), \( x \in \mathbb{R} \).
   
Given that \( f^{-1}(-3) = 4 \) and \( f(2) = 45 \), show that \( a = 2 \) and \( b = -8 \).

For these values of \( a \) and \( b \), find an expression for \( f^{-1} \).

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Let's go step by step to solve it.

### Step 1: Using \( f(2) = 45 \) to find a relation between \( a \) and \( b \)

We are given that \( f(x) = ax^2 + bx + 3 \) and \( f(2) = 45 \), meaning when \( x = 2 \), the output is 45. Substituting \( x = 2 \) into the function:

\[
f(2) = a(2)^2 + b(2) + 3 = 45
\]

This simplifies to:

\[
4a + 2b + 3 = 45
\]

Subtract 3 from both sides:

\[
4a + 2b = 42
\]

Now divide through by 2:

\[
2a + b = 21 \quad \text{(Equation 1)}
\]

### Step 2: Using \( f^{-1}(-3) = 4 \)

This means when \( f(4) = -3 \). So substitute \( x = 4 \) into the original function \( f(x) \) and set it equal to -3:

\[
f(4) = a(4)^2 + b(4) + 3 = -3
\]

This simplifies to:

\[
16a + 4b + 3 = -3
\]

Subtract 3 from both sides:

\[
16a + 4b = -6
\]

Now divide through by 2:

\[
8a + 2b = -3 \quad \text{(Equation 2)}
\]

### Step 3: Solving the system of equations

We now have the following system of equations:

1. \( 2a + b = 21 \)
2. \( 8a + 2b = -3 \)

Solve the first equation for \( b \):

\[
b = 21 - 2a
\]

Substitute this into the second equation:

\[
8a + 2(21 - 2a) = -3
\]

Simplify and solve for \( a \):

\[
8a + 42 - 4a = -3
\]

\[
4a = -3 - 42
\]

\[
4a = -45
\]

\[
a = -\frac{45}{4}


The function f(x) is defined by f(x) = ax^2 + bx + 3, where x ≥ 2, x ∈ ℝ. Given that f^-1(-3) = 4 and f(2) = 45, show that a = 2 and b = -8. For these values of a and b, find an expression for f^-1.

Learn how to find the values of a and b for a quadratic function f(x) = ax^2 + bx + 3, where f^-1(-3) = 4 and f(2) = 45. The solution involves solving systems of equations to derive a = 2 and b = -8, followed by finding the inverse function.

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