For the next two (2) items that follow:
\({\rm{f}}\left( {\rm{x}} \right) = \frac{{{{\rm{x}}^2} - 1}}{{{{\rm{x}}^2} + 1}}\) where x ϵ R At what value of x does f(x) attain minimum value?
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Finding the Minimum Value of the Function \(f(x)\)
The question asks us to find the value of \(x\) where the function \(f(x) = \frac{{{{\rm{x}}^2} - 1}}{{{{\rm{x}}^2} + 1}}\) attains its minimum value. The domain of the function is all real numbers, \(x \in \mathbb{R}\).
Understanding the Function \(f(x)\)
The function is a rational function involving \(x^2\). Notice that since \(x^2\) is always non-negative for real \(x\), the denominator \(x^2 + 1\) is always positive and at least 1. This means the function is defined for all real \(x\).
Method 1: Using Calculus (Derivative Test)
To find the minimum (or maximum) value of a differentiable function, we can find the critical points by computing the derivative \(f'(x)\) and setting it equal to zero.
We will use the quotient rule for differentiation, which states that if \(f(x) = \frac{u(x)}{v(x)}\), then \(f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}\).
In our case, let \(u(x) = x^2 - 1\) and \(v(x) = x^2 + 1\).
- Calculate the derivative of \(u(x)\): \(u'(x) = \frac{d}{dx}(x^2 - 1) = 2x\).
- Calculate the derivative of \(v(x)\): \(v'(x) = \frac{d}{dx}(x^2 + 1) = 2x\).
- The denominator squared is \((v(x))^2 = (x^2 + 1)^2\).
Now, substitute these into the quotient rule formula:
\(f'(x) = \frac{(2x)(x^2 + 1) - (x^2 - 1)(2x)}{(x^2 + 1)^2}\)
Expand the terms in the numerator:
\(f'(x) = \frac{2x^3 + 2x - (2x^3 - 2x)}{(x^2 + 1)^2}\)
\(f'(x) = \frac{2x^3 + 2x - 2x^3 + 2x}{(x^2 + 1)^2}\)
Simplify the numerator:
\(f'(x) = \frac{4x}{(x^2 + 1)^2}\)
Finding Critical Points
Critical points occur where \(f'(x) = 0\) or \(f'(x)\) is undefined. The denominator \((x^2 + 1)^2\) is always positive for real \(x\) (since \(x^2 \ge 0\), \(x^2+1 \ge 1\)), so \(f'(x)\) is never undefined.
Set the numerator equal to zero to find critical points:
\(4x = 0\)
\(x = 0\)
Thus, \(x = 0\) is the only critical point of the function.
Determining if it's a Minimum (First Derivative Test)
We can use the first derivative test to check if \(x = 0\) corresponds to a local minimum. We examine the sign of \(f'(x)\) in intervals around \(x = 0\).
- For \(x < 0\), let's pick \(x = -1\). \(f'(-1) = \frac{4(-1)}{((-1)^2 + 1)^2} = \frac{-4}{(1 + 1)^2} = \frac{-4}{4} = -1\). Since \(f'(-1) < 0\), \(f(x)\) is decreasing for \(x < 0\).
- For \(x > 0\), let's pick \(x = 1\). \(f'(1) = \frac{4(1)}{(1^2 + 1)^2} = \frac{4}{(1 + 1)^2} = \frac{4}{4} = 1\). Since \(f'(1) > 0\), \(f(x)\) is increasing for \(x > 0\).
Because the function changes from decreasing to increasing at \(x = 0\), this point is a local minimum. Since it's the only critical point and the function's behavior is consistent across the domain (decreasing then increasing), this local minimum is also the absolute minimum.
Method 2: Analyzing the Function's Structure
We can rewrite the function algebraically to understand its behavior:
\(f(x) = \frac{{{{\rm{x}}^2} - 1}}{{{{\rm{x}}^2} + 1}}\)
We can add and subtract 1 in the numerator:
\(f(x) = \frac{{{{\rm{x}}^2} + 1 - 2}}{{{{\rm{x}}^2} + 1}}\)
Separate the terms:
\(f(x) = \frac{{{{\rm{x}}^2} + 1}}{{{{\rm{x}}^2} + 1}} - \frac{2}{{{{\rm{x}}^2} + 1}}\)
\(f(x) = 1 - \frac{2}{{{{\rm{x}}^2} + 1}}\)
To find the minimum value of \(f(x)\), we need to minimize \(1 - \frac{2}{x^2 + 1}\). This expression is minimized when the term being subtracted, \(\frac{2}{x^2 + 1}\), is maximized. The fraction \(\frac{2}{x^2 + 1}\) is maximized when its denominator, \(x^2 + 1\), is minimized.
For any real number \(x\), \(x^2 \ge 0\). Therefore, \(x^2 + 1 \ge 0 + 1 = 1\). The minimum value of \(x^2 + 1\) is 1, which occurs when \(x^2 = 0\), meaning \(x = 0\).
When \(x = 0\), the denominator \(x^2 + 1\) is 1 (its minimum value). The term \(\frac{2}{x^2 + 1}\) becomes \(\frac{2}{1} = 2\) (its maximum value). The function value is then \(f(0) = 1 - 2 = -1\).
The minimum value of the function is -1, and this occurs at \(x = 0\).
Conclusion
Both the calculus method and the algebraic analysis show that the function \(f(x) = \frac{{{{\rm{x}}^2} - 1}}{{{{\rm{x}}^2} + 1}}\) reaches its minimum value at \(x = 0\).
| Value of x | Calculation of f(x) | Result | Observation |
|---|---|---|---|
| -1 (Option 1) | \(\frac{(-1)^2 - 1}{(-1)^2 + 1} = \frac{1 - 1}{1 + 1} = \frac{0}{2}\) | 0 | |
| 0 (Option 2) | \(\frac{(0)^2 - 1}{(0)^2 + 1} = \frac{0 - 1}{0 + 1} = \frac{-1}{1}\) | -1 | Minimum Value |
| 1 (Option 3) | \(\frac{(1)^2 - 1}{(1)^2 + 1} = \frac{1 - 1}{1 + 1} = \frac{0}{2}\) | 0 | |
| 2 (Option 4) | \(\frac{(2)^2 - 1}{(2)^2 + 1} = \frac{4 - 1}{4 + 1} = \frac{3}{5}\) | 0.6 |
The table evaluates the function at the given options, confirming that the minimum value among these occurs at \(x=0\).
Revision Table: Function Analysis and Minimum
| Concept | Description | Application in Finding Minimum |
|---|---|---|
| Function Minimum | The lowest value a function attains over its domain or a specific interval. | We seek the \(x\) value where \(f(x)\) is smallest. |
| Derivative | The rate of change of a function. For \(f(x)\), it's \(f'(x)\). | Used to locate critical points where the slope is zero. |
| Critical Point | A point \(x=c\) where \(f'(c)=0\) or \(f'(c)\) is undefined. | Potential location for a local minimum or maximum. |
| First Derivative Test | Examines the sign change of \(f'(x)\) around a critical point. | Confirms if a critical point is a minimum (sign changes from negative to positive). |
| Quotient Rule | A rule for finding the derivative of a fraction of two functions. | Necessary for calculating the derivative of the given rational function. |
Additional Information: Optimization Concepts
Finding the minimum or maximum of a function is a core concept in optimization. Optimization problems appear in various fields, like engineering, economics, and physics, where we want to find the best possible outcome (e.g., minimum cost, maximum profit, minimum energy).
For functions defined on an open interval or the entire real line, the absolute minimum (or maximum) often occurs at a critical point. If the domain is a closed interval, you must also evaluate the function at the endpoints of the interval.
In this specific problem, the function \(f(x) = 1 - \frac{2}{x^2 + 1}\) shows that as \(|x|\) becomes very large, \(x^2+1\) becomes very large, and \(\frac{2}{x^2+1}\) approaches 0. So, \(f(x)\) approaches \(1 - 0 = 1\) as \(|x| \to \infty\). Since the minimum value found is -1, which is less than 1, the minimum at \(x=0\) is indeed the absolute minimum.
Important Questions from Applications of Derivatives
The function is decreasing on :
The function attains local minimum value at :
What is the maximum value of y?
What is the maximum value of xy ?
Consider the following statements:
1. \({\rm{y}} = \frac{{{{\rm{e}}^{\rm{x}}} + {{\rm{e}}^{ - {\rm{x}}}}}}{2}\) is an increasing function on [0, ∞).
2. \({\rm{y}} = \frac{{{{\rm{e}}^{\rm{x}}} - {{\rm{e}}^{ - {\rm{x}}}}}}{2}\) is an increasing function on (-∞, ∞).
Which of the above statements is/are correct?
