To solve the problem, we need to prove that the function \( f(x) = 5x^2 + 6x - 9 \) is invertible, find its inverse, and then calculate \( f^{-1}(2) \) and \( f^{-1}(18) \).
### Step 1: Proving that \( f \) is invertible
To show that \( f \) is invertible, we need to prove that it is bijective, meaning it is both one-to-one (injective) and onto (surjective).
#### Step 1.1: Proving that \( f \) is one-to-one
To prove that \( f \) is one-to-one, we assume \( f(a) = f(b) \) for \( a, b \in R_+ \) and show that this implies \( a = b \).
1. Start with the equation:
\[
f(a) = f(b)
\]
This means:
\[
5a^2 + 6a - 9 = 5b^2 + 6b - 9
\]
2. Rearranging gives:
\[
5a^2 + 6a = 5b^2 + 6b
\]
\[
5a^2 - 5b^2 + 6a - 6b = 0
\]
3. Factor the left-hand side:
\[
5(a^2 - b^2) + 6(a - b) = 0
\]
\[
(a - b)(5(a + b) + 6) = 0
\]
4. Since \( a, b \in R_+ \), \( 5(a + b) + 6 > 0 \). Thus, \( a - b = 0 \) implies \( a = b \).
This shows that \( f \) is one-to-one.
#### Step 1.2: Proving that \( f \) is onto
To show that \( f \) is onto, we need to show that for every \( y \in [-9, \infty) \), there exists an \( x \in R_+ \) such that \( f(x) = y \).
1. Set \( y = f(x) \):
\[
y = 5x^2 + 6x - 9
\]
Rearranging gives:
\[
5x^2 + 6x - (y + 9) = 0
\]
2. This is a quadratic equation in \( x \). The discriminant must be non-negative for real solutions:
\[
D = b^2 - 4ac = 6^2 - 4 \cdot 5 \cdot (-y - 9) = 36 + 20y + 180 = 20y + 216
\]
Since \( y \geq -9 \), we have:
\[
D = 20y + 216 \geq 20(-9) + 216 = 36 \geq 0
\]
Thus, the quadratic equation has real solutions for all \( y \in [-9, \infty) \).
Since \( f \) is both one-to-one and onto, it is bijective, and therefore invertible.
### Step 2: Finding the inverse of \( f \)
We now find the inverse function \( f^{-1}(y) \).
1. From the equation:
\[
y = 5x^2 + 6x - 9
\]
Rearranging gives:
\[
5x^2 + 6x - (y + 9) = 0
\]
2. Using the quadratic formula \( x = \frac{-b \pm \sqrt{D}}{2a} \):
\[
x = \frac{-6 \pm \sqrt{20y + 216}}{10}
\]
Since \( x \) must be non-negative, we take the positive root:
\[
x = \frac{-6 + \sqrt{20y + 216}}{10}
\]
Therefore, the inverse function is:
\[
f^{-1}(y) = \frac{-6 + \sqrt{20y + 216}}{10}
\]
### Step 3: Finding \( f^{-1}(2) \) and \( f^{-1}(18) \)
1. **Calculating \( f^{-1}(2) \)**:
\[
f^{-1}(2) = \frac{-6 + \sqrt{20(2) + 216}}{10}
\]
\[
= \frac{-6 + \sqrt{40 + 216}}{10}
\]
\[
= \frac{-6 + \sqrt{256}}{10}
\]
\[
= \frac{-6 + 16}{10} = \frac{10}{10} = 1
\]
2. **Calculating \( f^{-1}(18) \)**:
\[
f^{-1}(18) = \frac{-6 + \sqrt{20(18) + 216}}{10}
\]
\[
= \frac{-6 + \sqrt{360 + 216}}{10}
\]
\[
= \frac{-6 + \sqrt{576}}{10}
\]
\[
= \frac{-6 + 24}{10} = \frac{18}{10} = \frac{9}{5}
\]
### Final Answers:
- \( f^{-1}(2) = 1 \)
- \( f^{-1}(18) = \frac{9}{5} \)
