Let `f : N rarr N` be a function defined as `f(x) = 4x^(2) + 12x + 15` is invertible (where S is range of f). Find the inverse of f and hence find `f^(-1)(31)`.

Answer

To find the inverse of the function \( f(x) = 4x^2 + 12x + 15 \) and then compute \( f^{-1}(31) \), we will follow these steps:

### Step 1: Set \( f(x) \) equal to \( y \)
Let \( y = f(x) = 4x^2 + 12x + 15 \).

### Step 2: Rearrange the equation to express \( x \) in terms of \( y \)
We need to rearrange the equation to solve for \( x \):
\[ y = 4x^2 + 12x + 15 \]
Subtract 15 from both sides:
\[ y - 15 = 4x^2 + 12x \]
Now, divide everything by 4:
\[ \frac{y - 15}{4} = x^2 + 3x \]

### Step 3: Rearrange into standard quadratic form
Rearranging gives us:
\[ x^2 + 3x - \frac{y - 15}{4} = 0 \]

### Step 4: Use the quadratic formula to solve for \( x \)
The quadratic formula is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
In our case, \( a = 1 \), \( b = 3 \), and \( c = -\frac{y - 15}{4} \). Plugging these values into the formula:
\[ x = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 1 \cdot \left(-\frac{y - 15}{4}\right)}}{2 \cdot 1} \]
This simplifies to:
\[ x = \frac{-3 \pm \sqrt{9 + (y - 15)}}{2} \]
\[ x = \frac{-3 \pm \sqrt{y - 6}}{2} \]

### Step 5: Choose the correct root
Since \( f(x) \) is a quadratic function that opens upwards, we take the positive root for the inverse:
\[ f^{-1}(y) = \frac{-3 + \sqrt{y - 6}}{2} \]

### Step 6: Find \( f^{-1}(31) \)
Now we compute \( f^{-1}(31) \):
\[ f^{-1}(31) = \frac{-3 + \sqrt{31 - 6}}{2} \]
Calculating inside the square root:
\[ f^{-1}(31) = \frac{-3 + \sqrt{25}}{2} \]
Since \( \sqrt{25} = 5 \):
\[ f^{-1}(31) = \frac{-3 + 5}{2} = \frac{2}{2} = 1 \]

### Final Answer
Thus, \( f^{-1}(31) = 1 \).