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In Exercises \(49-58\) , find the extreme values (absolute and local) of the function over its natural domain, and where they occur. $$ y=\frac{x}{x^{2}+1} $$

Short Answer

Expert verified
The absolute maximum is \( \frac{1}{2} \) at \( x = -1 \) and the absolute minimum is \( -\frac{1}{2} \) at \( x = 1 \).

Step by step solution

01

Find the derivative of the function

To find the extreme values, we first need to find the derivative of the function. The function given is \( y = \frac{x}{x^{2}+1} \). Let's use the quotient rule for differentiation, which is \( \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \). Here, \( u = x \) and \( v = x^2 + 1 \). Hence, \( \frac{du}{dx} = 1 \) and \( \frac{dv}{dx} = 2x \).
02

Apply the Quotient Rule

Using the quotient rule, the derivative is\[\frac{d}{dx}\left(\frac{x}{x^2+1}\right) = \frac{(x^2+1)(1) - x(2x)}{(x^2+1)^2} = \frac{x^2 + 1 - 2x^2}{(x^2+1)^2} = \frac{1-x^2}{(x^2+1)^2}.\]So, the derivative of the function is \( y' = \frac{1-x^2}{(x^2+1)^2} \).
03

Find Critical Points

To find the critical points, solve \( y' = 0 \). This gives\[\frac{1-x^2}{(x^2+1)^2} = 0.\]This means \( 1-x^2 = 0 \), or \( x^2 = 1 \). Hence, \( x = 1 \) or \( x = -1 \) are the critical points.
04

Evaluate the Function at Critical Points

Evaluate the function \( y = \frac{x}{x^2+1} \) at \( x = 1 \) and \( x = -1 \).- For \( x = 1 \): \[ y = \frac{1}{1^2+1} = \frac{1}{2}. \]- For \( x = -1 \): \[ y = \frac{-1}{(-1)^2+1} = \frac{-1}{2}. \]
05

Determine the Nature of Critical Points

Analyze the sign of the derivative in intervals around the critical points:- For \( x < -1 \), \( y' = \frac{1-x^2}{(x^2+1)^2} \) is positive since \( 1-x^2 \) is positive.- For \( -1 < x < 1 \), \( y' \) is negative since \( 1-x^2 \) is negative.- For \( x > 1 \), \( y' \) is positive again.Thus, \( x = -1 \) is a local maximum, and \( x = 1 \) is a local minimum.
06

Check Values at Infinity for Absolute Extremes

Consider the behavior of the function as \( x \to \pm \infty \):\[y = \frac{x}{x^2+1} \approx \frac{1}{x} \stackrel{x \to \pm \infty}{\longrightarrow} 0.\]Thus, the absolute maximum is \( \frac{1}{2} \) at \( x = -1 \), and the absolute minimum is \( -\frac{1}{2} \) at \( x = 1 \), since these are the highest and lowest values of \( y \) obtained.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Quotient Rule
The quotient rule is an essential tool in calculus when finding derivatives of functions that are expressed as a ratio or a fraction of two other functions. If you have a function expressed as \[ y = \frac{u(x)}{v(x)} \]where both the numerator \(u(x)\) and the denominator \(v(x)\) are functions of \(x\), then the quotient rule helps us compute its derivative. The rule states:\[ \left( \frac{u}{v} \right)' = \frac{v \cdot u' - u \cdot v'}{v^2} \]This formula provides a systematic way to handle derivatives of such functions, ensuring that no key terms are missed in the differentiation process.
- **Step 1**: Differentiate the numerator \(u(x)\) to get \(u'(x)\).
- **Step 2**: Differentiate the denominator \(v(x)\) to get \(v'(x)\).
- **Step 3**: Substitute these derivatives and the original functions \(u(x)\) and \(v(x)\) into the quotient rule formula.
Using this approach minimizes errors in more complex expressions and helps in finding critical points of functions, crucial in understanding their extreme values.
Critical Points
Critical points of a function give us insight into where the function's behavior changes, specifically in terms of local maxima or minima. To find these critical points, we first need the derivative of the original function set to zero. Critical points are the values of \(x\) where the derivative \(f'(x)\) is equal to zero or undefined.
- **Calculation**: Solve the equation \(f'(x) = 0\).
- **Interpretation**: These solutions \(x = a\) will potentially indicate a point where the slope of the tangent is zero, suggesting a local extremum.
In some cases, you may need to check if the derivative is also undefined at certain points, as these too can be critical points.
In the example \(y = \frac{x}{x^2 + 1}\), solving \(f'(x) = 0\) gave critical points at \(x = 1\) and \(x = -1\). These points are important in further steps of the analysis as they help evaluate the local extremum of the function.
Derivative Analysis
Derivative analysis is a powerful technique used to understand the behavior of functions across various intervals. It sheds light on the intervals where a function is increasing or decreasing, and more importantly, where it attains local or absolute extremums.
Analyzing the sign of the derivative involves:
  • Identifying critical points: where \(y' = 0\) or is undefined.
  • Testing intervals between critical points to determine where the derivative is positive (indicating the function is increasing) or negative (indicating it's decreasing).
For instance, looking at the derivative in different intervals around the critical points for the given function, you assess its behavior:
- For \(x < -1\): Positive derivative indicates increase.
- For \(-1 < x < 1\): Negative derivative indicates decrease.
- For \(x > 1\): Positive derivative indicates increase again.
This kind of analysis: - Helps confirm whether critical points are local maxima or minima.
- Shapes understanding of the function's complete behavior by using endpoints and infinity behavior to confirm absolute extremum.

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