To prove that the function \( f: [0, \infty) \to \mathbb{R} \) given by \( f(x) = 9x^2 + 6x - 5 \) is not invertible, we need to check if it is one-to-one (injective) and onto (surjective).
### Step 1: Check if the function is one-to-one
A function is one-to-one if \( f(x_1) = f(x_2) \) implies \( x_1 = x_2 \).
Assume \( f(x_1) = f(x_2) \):
\[
9x_1^2 + 6x_1 - 5 = 9x_2^2 + 6x_2 - 5
\]
This simplifies to:
\[
9x_1^2 + 6x_1 = 9x_2^2 + 6x_2
\]
Rearranging gives:
\[
9x_1^2 - 9x_2^2 + 6x_1 - 6x_2 = 0
\]
Factoring out gives:
\[
9(x_1^2 - x_2^2) + 6(x_1 - x_2) = 0
\]
Factoring further:
\[
(x_1 - x_2)(9(x_1 + x_2) + 6) = 0
\]
This means either \( x_1 - x_2 = 0 \) (which implies \( x_1 = x_2 \)) or \( 9(x_1 + x_2) + 6 = 0 \). The second equation does not yield valid solutions for \( x_1, x_2 \geq 0 \). Therefore, the function is one-to-one.
### Step 2: Check if the function is onto
To check if the function is onto, we need to see if every real number \( y \) in the codomain can be expressed as \( f(x) \) for some \( x \in [0, \infty) \).
The minimum value of \( f(x) \) occurs at the vertex of the parabola, given by:
\[
x = -\frac{b}{2a} = -\frac{6}{2 \cdot 9} = -\frac{1}{3}
\]
Since \( -\frac{1}{3} \) is not in the domain \( [0, \infty) \), we evaluate \( f(0) \):
\[
f(0) = 9(0)^2 + 6(0) - 5 = -5
\]
As \( x \to \infty \), \( f(x) \to \infty \). Therefore, the range of \( f \) is \( [-5, \infty) \).
Since the codomain is \( \mathbb{R} \) and the range is \( [-5, \infty) \), the function is not onto.
### Conclusion
Since \( f \) is not onto, it is not invertible.
### Step 3: Modify the codomain
To make \( f \) invertible, we can change the codomain to \( [-5, \infty) \).
### Step 4: Find the inverse function
To find the inverse, we set \( y = f(x) \):
\[
y = 9x^2 + 6x - 5
\]
Rearranging gives:
\[
9x^2 + 6x - (y + 5) = 0
\]
Using the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-6 \pm \sqrt{6^2 - 4 \cdot 9 \cdot (-y - 5)}}{2 \cdot 9}
\]
Calculating the discriminant:
\[
= \frac{-6 \pm \sqrt{36 + 36y + 180}}{18} = \frac{-6 \pm \sqrt{36y + 216}}{18}
\]
Simplifying:
\[
= \frac{-6 \pm 6\sqrt{y + 6}}{18} = \frac{-1 \pm \sqrt{y + 6}}{3}
\]
Since \( x \) must be non-negative, we take the positive root:
\[
f^{-1}(y) = \frac{-1 + \sqrt{y + 6}}{3}
\]
### Final Answer
Thus, the inverse function is:
\[
f^{-1}(y) = \frac{-1 + \sqrt{y + 6}}{3}, \quad \text{for } y \in [-5, \infty)
\]
