Let `f:DtoR`, where D is the domain of `f`. Find the inverse of `f` if it exists:
Let `f:[0,3]to[1,13]` is defined by `f(x)=x^(2)+x+1`, then find `f^(-1)(x)`.

Explanation

To find the inverse of the function \( f(x) = x^2 + x + 1 \) defined on the interval \([0, 3]\) with the range \([1, 13]\), we will follow these steps:

### Step 1: Verify if the function is one-to-one
To check if the function is one-to-one, we need to find its derivative and see if it is always positive (indicating the function is increasing).

**Derivative Calculation:**
\[ f'(x) = 2x + 1 \]
For \( x \) in the interval \([0, 3]\):
- At \( x = 0 \): \( f'(0) = 2(0) + 1 = 1 \) (positive)
- At \( x = 3 \): \( f'(3) = 2(3) + 1 = 7 \) (positive)

Since \( f'(x) > 0 \) for all \( x \) in \([0, 3]\), \( f(x) \) is strictly increasing and thus one-to-one.

### Step 2: Verify if the function is onto
Next, we check if the function covers the entire range \([1, 13]\).

**Function Values at Endpoints:**
- \( f(0) = 0^2 + 0 + 1 = 1 \)
- \( f(3) = 3^2 + 3 + 1 = 9 + 3 + 1 = 13 \)

Since \( f(0) = 1 \) and \( f(3) = 13 \), and the function is continuous on \([0, 3]\), it covers all values from \( 1 \) to \( 13 \). Thus, it is onto.

### Step 3: Find the inverse function
Since the function is both one-to-one and onto, we can find the inverse. We start by setting \( y = f(x) \):
\[ y = x^2 + x + 1 \]
Now, we need to solve for \( x \) in terms of \( y \):
\[ x^2 + x + (1 - y) = 0 \]
This is a quadratic equation in \( x \). We can use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1 \), \( b = 1 \), and \( c = 1 - y \):
\[ x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (1 - y)}}{2 \cdot 1} \]
\[ x = \frac{-1 \pm \sqrt{1 - 4 + 4y}}{2} \]
\[ x = \frac{-1 \pm \sqrt{4y - 3}}{2} \]
Since \( f(x) \) is increasing, we take the positive root:
\[ x = \frac{-1 + \sqrt{4y - 3}}{2} \]
Thus, the inverse function is:
\[ f^{-1}(y) = \frac{-1 + \sqrt{4y - 3}}{2} \]

### Final Answer
The inverse function is:
\[ f^{-1}(x) = \frac{-1 + \sqrt{4x - 3}}{2} \]