Show that the function f:A $$ \rarr $$R defined by f​$$ \left(x\right) $$=5$$ -\frac{1}{\left(1+x\right)},A= $${$$ x\in R $$|$$ x\ne-1 $$}, is one-one. Find the range and the suitable inverse

To show that the function $ f: A \to \mathbb{R} $ defined by $ f(x) = 5 - \frac{1}{1+x} $ for $ A = \{ x \in \mathbb{R} \mid x \neq -1 \} $ is one-one, and to find its range and inverse, we proceed as follows:

### Proving that $ f(x) $ is One-One

A function is one-one (injective) if $ f(x_1) = f(x_2) $ implies $ x_1 = x_2 $. We start by assuming:

\[ f(x_1) = f(x_2) \]

Substituting the definition of $ f $:

\[ 5 - \frac{1}{1+x_1} = 5 - \frac{1}{1+x_2} \]

Subtracting 5 from both sides:

\[ - \frac{1}{1+x_1} = - \frac{1}{1+x_2} \]

Multiplying both sides by -1:

\[ \frac{1}{1+x_1} = \frac{1}{1+x_2} \]

Taking the reciprocals of both sides:

\[ 1+x_1 = 1+x_2 \]

Subtracting 1 from both sides:

\[ x_1 = x_2 \]

Since $ f(x_1) = f(x_2) $ implies $ x_1 = x_2 $, the function $ f $ is injective.

### Finding the Range of $ f $

To find the range, we consider the expression $ f(x) = 5 - \frac{1}{1+x} $. We examine the behavior of $ f $ as $ x $ approaches various values:

1. As $ x \to \infty $, $ \frac{1}{1+x} \to 0 $ and hence $ f(x) \to 5 $.
2. As $ x \to -1^+ $, $ \frac{1}{1+x} \to \infty $ and hence $ f(x) \to -\infty $.
3. As $ x \to -1^- $, $ \frac{1}{1+x} \to -\infty $ and hence $ f(x) \to \infty $.

The function $ f(x) $ can take any real value because $ f $ is continuous and covers all values between $ -\infty $ and $ 5 $, and $ 5 $ itself is approached as $ x \to \infty $. Therefore, the range of $ f $ is:

\[ \text{Range of } f = (-\infty, 5) \]

### Finding the Inverse of $ f $

To find the inverse, we start with the equation $ y = 5 - \frac{1}{1+x} $ and solve for $ x $:

\[ y = 5 - \frac{1}{1+x} \]

Subtract 5 from both sides:

\[ y - 5 = - \frac{1}{1+x} \]

Multiply both sides by -1:

\[ 5 - y = \frac{1}{1+x} \]

Take the reciprocal of both sides:

\[ \frac{1}{5 - y} = 1 + x \]

Subtract 1 from both sides:

\[ x = \frac{1}{5 - y} - 1 \]

Therefore, the inverse function $ f^{-1} $ is:

\[ f^{-1}(y) = \frac{1}{5 - y} - 1 \]

### Summary

1. The function $ f(x) = 5 - \frac{1}{1+x} $ is injective.
2. The range of $ f $ is $ (-\infty, 5) $.
3. The inverse function $ f^{-1}(y) $ is $ \frac{1}{5 - y} - 1 $.

Would you like further details on any of these steps or have any questions?

### Relative Questions
1. What does it mean for a function to be injective (one-one)?
2. How can we verify if a function is surjective (onto)?
3. What is the significance of finding the range of a function?
4. How do we derive the inverse of a function?
5. Can you explain the behavior of the function as $ x \to -1 $?
6. Why is the function not defined at $ x = -1 $?
7. What is the graphical interpretation of an inverse function?
8. How does the concept of continuity help in determining the range?

### Tip
When finding the inverse of a function, always ensure that the function is one-one and onto within the specified domain and range to guarantee that the inverse function exists and is well-defined.

Learn how to prove that the function f(x) = 5 - 1/(1+x) is one-one (injective) and find its range and inverse function. This detailed explanation covers the concept of injectivity in functions and the process of finding inverse functions. Suitable for undergraduate level study.

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