If `f(x)=(x^(2)-x)/(x^(2)+2x)`, then find the domain and range of f. Show that f is one-one. Also, find the function `(d(f^(-1)(x)))/(dx)` and its domain.

Explanation

To solve the problem step by step, we will find the domain and range of the function \( f(x) = \frac{x^2 - x}{x^2 + 2x} \), show that it is one-one, and find the derivative of its inverse function along with its domain.

### Step 1: Finding the Domain of \( f(x) \)

The function \( f(x) \) is defined as long as the denominator is not equal to zero. We need to solve the equation:

\[ x^2 + 2x \neq 0 \]

Factoring the denominator gives:

\[ x(x + 2) \neq 0 \]

This implies:

\[ x \neq 0 \quad \text{and} \quad x \neq -2 \]

Thus, the domain of \( f \) is all real numbers except 0 and -2:

\[ \text{Domain} = \mathbb{R} \setminus \{0, -2\} \]

### Step 2: Finding the Range of \( f(x) \)

To find the range, we will analyze the behavior of \( f(x) \). We can rewrite \( f(x) \):

\[ f(x) = \frac{x(x - 1)}{x(x + 2)} = \frac{x - 1}{x + 2} \quad \text{(for } x \neq 0\text{)} \]

Next, we will find the horizontal asymptote by evaluating the limits as \( x \) approaches infinity:

\[ \lim_{x \to \infty} f(x) = \lim_{x \to \infty} \frac{x - 1}{x + 2} = 1 \]
\[ \lim_{x \to -\infty} f(x) = \lim_{x \to -\infty} \frac{x - 1}{x + 2} = 1 \]

Now, we check for any horizontal asymptotes or values that \( f(x) \) cannot take. Setting \( f(x) = k \):

\[ \frac{x - 1}{x + 2} = k \implies x - 1 = k(x + 2) \implies x - 1 = kx + 2k \]
\[ x - kx = 2k + 1 \implies x(1 - k) = 2k + 1 \implies x = \frac{2k + 1}{1 - k} \quad \text{(provided } k \neq 1\text{)} \]

Thus, \( f(x) \) can take all real values except for \( k = 1 \). Therefore, the range of \( f \) is:

\[ \text{Range} = \mathbb{R} \setminus \{1\} \]

### Step 3: Showing that \( f \) is One-One

To show that \( f \) is one-one, we assume \( f(a) = f(b) \):

\[ \frac{a^2 - a}{a^2 + 2a} = \frac{b^2 - b}{b^2 + 2b} \]

Cross-multiplying gives:

\[ (a^2 - a)(b^2 + 2b) = (b^2 - b)(a^2 + 2a) \]

Expanding both sides:

\[ a^2b^2 + 2a^2b - ab^2 - 2ab = b^2a^2 + 2b^2a - ba^2 - 2ba \]

Cancelling \( a^2b^2 \) from both sides:

\[ 2a^2b - ab^2 - 2ab = 2b^2a - ba^2 - 2ba \]

Rearranging gives:

\[ 2a^2b - 2b^2a = ab^2 - ba^2 \]

Factoring out common terms:

\[ 2ab(a - b) = ab(b - a) \]

Since \( ab \neq 0 \) (as \( a \) and \( b \) cannot be 0 or -2), we can divide both sides by \( ab \):

\[ 2(a - b) = -(b - a) \implies 3(a - b) = 0 \implies a = b \]

Thus, \( f \) is one-one.

### Step 4: Finding the Derivative of the Inverse Function

To find \( \frac{d(f^{-1}(x))}{dx} \), we use the formula:

\[ \frac{d(f^{-1}(x))}{dx} = \frac{1}{f'(f^{-1}(x))} \]

First, we need to find \( f'(x) \):

Using the quotient rule:

\[ f'(x) = \frac{(x^2 + 2x)(2x - 1) - (x^2 - x)(2x + 2)}{(x^2 + 2x)^2} \]

Calculating the numerator:

\[ = (2x^3 + 4x^2 - 2x^2 - 2x) - (2x^3 + 2x^2 - 2x^2 + 2x) \]
\[ = 2x^3 + 4x^2 - 2x^2 - 2x - 2x^3 - 2x^2 + 2x = 2x^2 \]

Thus,

\[ f'(x) = \frac{2x^2}{(x^2 + 2x)^2} \]

Now, substituting into the inverse derivative formula:

\[ \frac{d(f^{-1}(x))}{dx} = \frac{(x^2 + 2x)^2}{2x^2} \]

### Step 5: Domain of \( \frac{d(f^{-1}(x))}{dx} \)

The domain of \( \frac{d(f^{-1}(x))}{dx} \) is determined by the range of \( f(x) \) since \( f^{-1} \) exists only for values in the range of \( f \):

\[ \text{Domain} = \mathbb{R} \setminus \{1\} \]

### Summary of Results

- **Domain of \( f \)**: \( \mathbb{R} \setminus \{0, -2\} \)
- **Range of \( f \)**: \( \mathbb{R} \setminus \{1\} \)
- **\( f \) is one-one**: Yes
- **Derivative of the inverse**: \( \frac{d(f^{-1}(x))}{dx} = \frac{(x^2 + 2x)^2}{2x^2} \)
- **Domain of \( \frac{d(f^{-1}(x))}{dx} \)**: \( \mathbb{R} \setminus \{1\} \)