Question

Let \(f: R-\\{1,2,3\\} \rightarrow R\) be a function defined by \(f(x)=\frac{1}{x-1}+\frac{2}{x-2}+\frac{3}{x-3} .\) Then \(f(x)\) is (1) one-one (2) many one (3) monotonocally increasing function (4) monotonocally decreasing function

Short answer

The function \(f(x)\) is a monotonically decreasing function (Option 4).

Step-by-step solution

- Find the domain of the function

The function is defined as \(f(x) = \frac{1}{x-1} + \frac{2}{x-2} + \frac{3}{x-3}\). The domain consists of all real numbers except for \(x=1\), \(x=2\), and \(x=3\) because these values make the denominators zero and the function undefined.

- Take the derivative of the function

To understand the behavior of the function, compute the derivative \(f'(x)\). \[ f'(x) = -\frac{1}{(x-1)^2} - \frac{2}{(x-2)^2} - \frac{3}{(x-3)^2} \]

- Analyze the sign of the derivative

Examine the sign of the derivative \(f'(x)\). Each term in the derivative \(f'(x)\) has a negative sign and the squares in the denominators are always positive. Thus, all terms are negative, making \(f'(x) < 0\) for all values in the domain.

- Determine monotonicity

Since the derivative is always negative, the function \(f(x)\) is monotonically decreasing.

- Conclusion

From the analysis, we conclude that among the given choices, the function \(f(x)\) is a monotonically decreasing function. The correct choice is (4).

Key concepts

Monotonically Decreasing Function

A monotonically decreasing function is one where the function's value never increases as the input increases. In simpler words, as you move from left to right along the x-axis, the function's output decreases or stays constant.
To determine if a function is monotonically decreasing, we often use its derivative. If the derivative is always negative over the function's domain, the function is monotonically decreasing.
In our exercise, the function is defined as:\[ f(x) = \frac{1}{x-1} + \frac{2}{x-2} + \frac{3}{x-3} \]We found its derivative:\[ f'(x) = -\frac{1}{(x-1)^2} - \frac{2}{(x-2)^2} - \frac{3}{(x-3)^2} \]Since each term in the derivative is negative and the denominators are squared (making them always positive), the entire derivative is always negative. Therefore, the function is confirmed to be monotonically decreasing across its domain.

Function Domain

The function domain refers to all possible input values (x-values) for which the function is defined.
For the given function:\[ f(x) = \frac{1}{x-1} + \frac{2}{x-2} + \frac{3}{x-3} \]It is clear that the function has restrictions. Specifically, the function becomes undefined if any denominator equals zero (because division by zero is undefined). Hence, we need to exclude the values of x that make any denominator zero.
Here, those values are:

• x = 1
• x = 2
• x = 3

Therefore, the domain of the given function is all real numbers except for 1, 2, and 3. In set notation, this is written as:\[ \text{Domain} = \text{R} - \{1, 2, 3\} \]

Derivative Analysis

The derivative of a function is a tool from calculus that helps us understand the rate of change of the function. It provides insights into the function's behavior, such as where it is increasing or decreasing.
For our given function, we calculated the derivative as:\[ f'(x) = -\frac{1}{(x-1)^2} - \frac{2}{(x-2)^2} - \frac{3}{(x-3)^2} \]By analyzing this derivative, we saw that each term in the expression is a negative value due to the negative signs in the numerators and the fact that squared terms in the denominators are always positive.
For each value of x in the function's domain, the derivative is negative, indicating that the function's rate of change is always negative. This behavior confirms that the function is monotonically decreasing throughout its domain. In summary:

• If the derivative is positive, the function is increasing.
• If the derivative is negative, the function is decreasing.
• If the derivative is zero, the function is constant (at least locally).