Prove that the function f : [`0,oo)rarrR`, given by `f(x)=9x^(2)+6x-5` is not invertible.
Modify the co-domain of the function f to make it invertible, and hence, find `f^(-1)`.

Answer

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 a one-to-one (injective) and onto (surjective) function.

### Step 1: Check if the function is one-to-one (injective)

A function is one-to-one if it is either always increasing or always decreasing in its domain. To check this, we will find the derivative of the function.

1. **Find the derivative**:
\[ f'(x) = \frac{d}{dx}(9x^2 + 6x - 5) = 18x + 6 \]

2. **Analyze the derivative**:
The derivative \( f'(x) = 18x + 6 \) is always positive for \( x \geq 0 \) because:
- At \( x = 0 \), \( f'(0) = 6 > 0 \)
- For \( x > 0 \), \( f'(x) > 6 > 0 \)

Since \( f'(x) > 0 \) for all \( x \in [0, \infty) \), the function \( f(x) \) is strictly increasing on its domain.

### Step 2: Check if the function is onto (surjective)

To determine if the function is onto, we need to compare the range of the function with its codomain.

1. **Find the minimum value of the function**:
Since \( f(x) \) is a quadratic function that opens upwards (the coefficient of \( x^2 \) is positive), the minimum value occurs at the vertex. The vertex \( x \) can be found using the formula \( x = -\frac{b}{2a} \):
\[ x = -\frac{6}{2 \cdot 9} = -\frac{1}{3} \]
However, since our domain is \( [0, \infty) \), we evaluate \( f(0) \):
\[ f(0) = 9(0)^2 + 6(0) - 5 = -5 \]

2. **Determine the range**:
As \( x \to \infty \), \( f(x) \to \infty \). Therefore, the range of \( f \) is \( [-5, \infty) \).

3. **Compare with the codomain**:
The codomain is \( \mathbb{R} \) (all real numbers), while the range is \( [-5, \infty) \). Since the range does not cover all real numbers, the function is not onto.

### Conclusion:
Since \( f \) is not onto, it is not invertible.

### Step 3: Modify the codomain to make it invertible

To make the function invertible, we can redefine the codomain to be equal to the range of the function:
\[ \text{New codomain} = [-5, \infty) \]

### Step 4: Find the inverse of the function

To find the inverse, we will replace \( f(x) \) with \( y \) and solve for \( x \):
\[ y = 9x^2 + 6x - 5 \]
Rearranging gives:
\[ 9x^2 + 6x - (y + 5) = 0 \]

Now, we can use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here, \( a = 9 \), \( b = 6 \), and \( c = -(y + 5) \).

Calculating the discriminant:
\[ D = b^2 - 4ac = 6^2 - 4 \cdot 9 \cdot (-(y + 5)) = 36 + 36(y + 5) = 36y + 216 \]

Now applying the quadratic formula:
\[ x = \frac{-6 \pm \sqrt{36y + 216}}{18} \]
\[ x = \frac{-6 \pm 6\sqrt{y + 6}}{18} \]
\[ x = \frac{-1 \pm \sqrt{y + 6}}{3} \]

Since \( x \) must be non-negative (as per the domain), we take the positive root:
\[ x = \frac{-1 + \sqrt{y + 6}}{3} \]

Thus, the inverse function is:
\[ f^{-1}(y) = \frac{-1 + \sqrt{y + 6}}{3} \]

### Final Answer:
The function \( f(x) = 9x^2 + 6x - 5 \) is not invertible in its original form. By modifying the codomain to \( [-5, \infty) \), we make it invertible, and the inverse function is:
\[ f^{-1}(y) = \frac{-1 + \sqrt{y + 6}}{3} \]