Analyzing Function Behavior: \({\rm{f}}\left( {\rm{x}} \right) = {\rm{x}} + \frac{1}{{\rm{x}}}\) on (0, 1)

The question asks about the behavior of the function \({\rm{f}}\left( {\rm{x}} \right) = {\rm{x}} + \frac{1}{{\rm{x}}}\) within the specific interval \({\rm{x}} \in\) (0, 1).

To determine if a function is increasing or decreasing in an interval, we typically analyze the sign of its first derivative in that interval. If the first derivative is positive, the function is increasing. If it's negative, the function is decreasing. If it changes sign within the interval, the function might fluctuate.

Finding the First Derivative

Let's find the first derivative of \({\rm{f}}\left( {\rm{x}} \right)\):

Given \({\rm{f}}\left( {\rm{x}} \right) = {\rm{x}} + \frac{1}{{\rm{x}}}\). We can rewrite \(\frac{1}{{\rm{x}}}\) as \({\rm{x}}^{ - 1}\).

So, \({\rm{f}}\left( {\rm{x}} \right) = {\rm{x}} + {\rm{x}}^{ - 1}\).

Now, we differentiate \({\rm{f}}\left( {\rm{x}} \right)\) with respect to \({\rm{x}}\):

\({\rm{f}}'\left( {\rm{x}} \right) = \frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{\rm{x}} + {\rm{x}}^{ - 1}} \right)\)

\({\rm{f}}'\left( {\rm{x}} \right) = \frac{{\rm{d}}}{{{\rm{dx}}}}\left( {\rm{x}} \right) + \frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{\rm{x}}^{ - 1}} \right)\)

Using the power rule \(\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{\rm{x}}^{\rm{n}}} \right) = {\rm{nx}}^{{\rm{n}} - 1}\):

• \(\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {\rm{x}} \right) = \frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{\rm{x}}^1} \right) = 1 \cdot {\rm{x}}^{1 - 1} = 1 \cdot {\rm{x}}^0 = 1 \cdot 1 = 1\)
• \(\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{\rm{x}}^{ - 1}} \right) = -1 \cdot {\rm{x}}^{ - 1 - 1} = -1 \cdot {\rm{x}}^{ - 2} = - \frac{1}{{{\rm{x}}^2}}\)

So, the first derivative is \({\rm{f}}'\left( {\rm{x}} \right) = 1 - \frac{1}{{{\rm{x}}^2}}\).

Analyzing the Derivative in the Interval (0, 1)

We need to determine the sign of \({\rm{f}}'\left( {\rm{x}} \right) = 1 - \frac{1}{{{\rm{x}}^2}}\) for \({\rm{x}} \in\) (0, 1).

Consider any value of \({\rm{x}}\) such that \(0 < {\rm{x}} < 1\).

If \(0 < {\rm{x}} < 1\), then squaring \({\rm{x}}\) will result in an even smaller positive number. So, \(0 < {\rm{x}}^2 < 1\).

Now, let's look at the term \(\frac{1}{{{\rm{x}}^2}}\). If a positive number is less than 1, its reciprocal is greater than 1.

Since \(0 < {\rm{x}}^2 < 1\), it follows that \(\frac{1}{{{\rm{x}}^2}} > 1\).

Now substitute this back into the expression for \({\rm{f}}'\left( {\rm{x}} \right)\):

\({\rm{f}}'\left( {\rm{x}} \right) = 1 - \frac{1}{{{\rm{x}}^2}}\)

Since \(\frac{1}{{{\rm{x}}^2}} > 1\), when we subtract a number greater than 1 from 1, the result will be negative.

For example, if \({\rm{x}} = 0.5\), then \({\rm{x}}^2 = 0.25\), and \(\frac{1}{{{\rm{x}}^2}} = \frac{1}{0.25} = 4\). Then \({\rm{f}}'\left( {0.5} \right) = 1 - 4 = -3\), which is negative.

Thus, for all \({\rm{x}} \in\) (0, 1), \({\rm{f}}'\left( {\rm{x}} \right) < 0\).

Conclusion on Function Behavior

Because the first derivative \({\rm{f}}'\left( {\rm{x}} \right)\) is negative for all \({\rm{x}}\) in the interval (0, 1), the function \({\rm{f}}\left( {\rm{x}} \right)\) is decreasing in this interval.

Matching with Options

Let's compare our finding with the given options:

• Option 1: f(x) fluctuates in the interval. This is incorrect because the derivative has a consistent sign.
• Option 2: f(x) increases in the interval. This is incorrect because the derivative is negative.
• Option 3: f(x) decreases in the interval. This matches our conclusion.
• Option 4: None of the above. This is incorrect as option 3 is correct.

Therefore, the correct statement is that \({\rm{f}}\left( {\rm{x}} \right)\) decreases in the interval (0, 1).

Revision Table: Function Behavior Analysis

Additional Information: Increasing and Decreasing Functions

The behavior of a function (whether it's increasing, decreasing, or constant) over an interval can be formally defined using calculus:

• Increasing Function: A function \({\rm{f}}\left( {\rm{x}} \right)\) is increasing on an interval (a, b) if for any two numbers \({\rm{x}}_1\) and \({\rm{x}}_2\) in (a, b) with \({\rm{x}}_1 < {\rm{x}}_2\), we have \({\rm{f}}\left( {{\rm{x}}_1} \right) < {\rm{f}}\left( {{\rm{x}}_2} \right)\). Using calculus, if \({\rm{f}}'\left( {\rm{x}} \right) > 0\) for all \({\rm{x}} \in\) (a, b), then \({\rm{f}}\left( {\rm{x}} \right)\) is increasing on (a, b).
• Decreasing Function: A function \({\rm{f}}\left( {\rm{x}} \right)\) is decreasing on an interval (a, b) if for any two numbers \({\rm{x}}_1\) and \({\rm{x}}_2\) in (a, b) with \({\rm{x}}_1 < {\rm{x}}_2\), we have \({\rm{f}}\left( {{\rm{x}}_1} \right) > {\rm{f}}\left( {{\rm{x}}_2} \right)\). Using calculus, if \({\rm{f}}'\left( {\rm{x}} \right) < 0\) for all \({\rm{x}} \in\) (a, b), then \({\rm{f}}\left( {\rm{x}} \right)\) is decreasing on (a, b).
• Constant Function: A function \({\rm{f}}\left( {\rm{x}} \right)\) is constant on an interval (a, b) if for any two numbers \({\rm{x}}_1\) and \({\rm{x}}_2\) in (a, b), we have \({\rm{f}}\left( {{\rm{x}}_1} \right) = {\rm{f}}\left( {{\rm{x}}_2} \right)\). Using calculus, if \({\rm{f}}'\left( {\rm{x}} \right) = 0\) for all \({\rm{x}} \in\) (a, b), then \({\rm{f}}\left( {\rm{x}} \right)\) is constant on (a, b).

In this specific problem, \({\rm{f}}'\left( {\rm{x}} \right)\) was found to be strictly negative throughout the interval (0, 1), confirming the decreasing nature of the function on this interval.