Let `f:N rarr R` be a function defined as `f(x)=4x^(2)+12x+15`.
Show that `f:N rarr S`, where S is the range of f, is invertible. Also find the inverse of f. Hence find `f^(-1)(31)`.

Answer

To solve the problem step by step, we will follow the instructions given in the question.

### Step 1: Define the function
We have the function \( f: \mathbb{N} \to \mathbb{R} \) defined as:
\[ f(x) = 4x^2 + 12x + 15 \]

### Step 2: Simplify the function
We can rewrite the function in a simpler form:
\[ f(x) = 4(x^2 + 3x) + 15 = 4\left(x + \frac{3}{2}\right)^2 + 6 \]
This shows that the function is a quadratic function that opens upwards.

### Step 3: Show that \( f \) is one-to-one (injective)
To show that \( f \) is one-to-one, we assume \( f(a) = f(b) \) for \( a, b \in \mathbb{N} \):
\[ 4a^2 + 12a + 15 = 4b^2 + 12b + 15 \]
Subtracting \( 15 \) from both sides:
\[ 4a^2 + 12a = 4b^2 + 12b \]
Dividing by \( 4 \):
\[ a^2 + 3a = b^2 + 3b \]
Rearranging gives:
\[ a^2 - b^2 + 3a - 3b = 0 \]
Factoring:
\[ (a - b)(a + b) + 3(a - b) = 0 \]
Factoring out \( (a - b) \):
\[ (a - b)(a + b + 3) = 0 \]
This implies either \( a - b = 0 \) (hence \( a = b \)) or \( a + b + 3 = 0 \) (which is impossible for natural numbers). Thus, \( f \) is one-to-one.

### Step 4: Show that \( f \) is onto (surjective)
To show that \( f \) is onto, we need to show that for every \( y \in S \) (the range of \( f \)), there exists an \( x \in \mathbb{N} \) such that \( f(x) = y \).

From the simplified form:
\[ f(x) = 4\left(x + \frac{3}{2}\right)^2 + 6 \]
The minimum value occurs when \( x = 0 \):
\[ f(0) = 4\left(0 + \frac{3}{2}\right)^2 + 6 = 4 \cdot \frac{9}{4} + 6 = 9 + 6 = 15 \]
Thus, the range \( S \) starts from \( 15 \) and goes to infinity. For any \( y \geq 15 \), we can find \( x \) such that:
\[ y = 4\left(x + \frac{3}{2}\right)^2 + 6 \]
Rearranging gives:
\[ \left(x + \frac{3}{2}\right)^2 = \frac{y - 6}{4} \]
Taking square roots:
\[ x + \frac{3}{2} = \sqrt{\frac{y - 6}{4}} \quad \text{or} \quad x + \frac{3}{2} = -\sqrt{\frac{y - 6}{4}} \quad (\text{not valid since } x \in \mathbb{N}) \]
Thus:
\[ x = \sqrt{\frac{y - 6}{4}} - \frac{3}{2} \]
This shows that for every \( y \geq 15 \), there exists an \( x \in \mathbb{N} \), proving \( f \) is onto.

### Step 5: Conclusion on invertibility
Since \( f \) is both one-to-one and onto, it is invertible.

### Step 6: Find the inverse function
From the previous calculations, we have:
\[ f^{-1}(y) = \sqrt{\frac{y - 6}{4}} - \frac{3}{2} \]
Replacing \( y \) with \( x \):
\[ f^{-1}(x) = \sqrt{\frac{x - 6}{4}} - \frac{3}{2} \]

### Step 7: Find \( f^{-1}(31) \)
Now we calculate:
\[ f^{-1}(31) = \sqrt{\frac{31 - 6}{4}} - \frac{3}{2} = \sqrt{\frac{25}{4}} - \frac{3}{2} = \frac{5}{2} - \frac{3}{2} = 1 \]

Thus, the final answer is:
\[ f^{-1}(31) = 1 \]