Problem 57 - Thomas Calculus in SI Units

Question

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

Step-by-Step Solution

Answer

Absolute maximum at \( y = \frac{1}{2} \) (\( x = 1 \)); minimum at \( y = -\frac{1}{2} \) (\( x = -1 \)).

Step 1: Find the Critical Points

To find the extreme values, we first need to determine the critical points of the function. We do this by finding the derivative and setting it equal to zero. The function is given by \[ y = \frac{x}{x^{2}+1} \] Compute the derivative using the quotient rule. Let \( u = x \) and \( v = x^2 + 1 \), then:\[ y' = \frac{v \cdot u' - u \cdot v'}{v^2} = \frac{(x^2 + 1) \cdot 1 - x \cdot 2x}{(x^2 + 1)^2} = \frac{x^2 + 1 - 2x^2}{(x^2 + 1)^2} = \frac{1 - x^2}{(x^2 + 1)^2} \] Set the derivative equal to zero to find critical points: \[ \frac{1 - x^2}{(x^2 + 1)^2} = 0 \] This implies: \[ 1 - x^2 = 0 \] Solve for \( x \): \[ x^2 = 1 \] \[ x = \pm 1 \] Now we have two critical points: \( x = 1 \) and \( x = -1 \).

Step 2: Determine the Second Derivative

We will use the second derivative test to classify the critical points as local maxima or minima. Compute the second derivative of the function:Start with the first derivative:\[ y' = \frac{1 - x^2}{(x^2 + 1)^2} \] Differentiate again using quotient rule and chain rule:Let \( u = 1 - x^2 \), \( v = (x^2 + 1)^2 \):\[ y'' = \frac{v \cdot u' - u \cdot v'}{v^2} \]Finding derivatives:- \( u' = -2x \)- \( v' = 2(x^2 + 1)\cdot 2x = 4x(x^2 + 1) \)Substitute these:\[ y'' = \frac{(x^2 + 1)^2 (-2x) - (1 - x^2)(4x(x^2 + 1))}{(x^2 + 1)^4} \]Simplifying:\[ y'' = \frac{-2x(x^2 + 1)^2 - 4x(x^2 - 1)(x^2 + 1)}{(x^2 + 1)^4} = \frac{-2x((x^2 + 1)^2 + 2(x^2 - 1)(x^2 + 1))}{(x^2 + 1)^4} \]Evaluate \( y'' \) at the critical points: - For \( x = 1 \): \[ y''(1) \approx \text{negative value} \] Hence, \( x = 1 \) is a local maximum.- For \( x = -1 \): \[ y''(-1) \approx \text{positive value} \] Hence, \( x = -1 \) is a local minimum.

Step 3: Determine Absolute Extrema

To find absolute extrema over the entire domain, we must consider the behavior of \( y \) as \( x \to \infty \) and \( x \to -\infty \).The function is\[ y = \frac{x}{x^2 + 1} \]Consider:- As \( x \to \infty \), \( y \to 0 \).- As \( x \to -\infty \), \( y \to 0 \).Check function values at critical points:- \( y(1) = \frac{1}{2} \), maximum value.- \( y(-1) = -\frac{1}{2} \), minimum value.Therefore, the absolute maximum is \( \frac{1}{2} \) at \( x = 1 \), and the absolute minimum is \( -\frac{1}{2} \) at \( x = -1 \).

Key Concepts

Critical Points

Critical points are special places on the graph of a function where the function's rate of change stops, which often means the slope of the tangent is zero or undefined. Finding these points is an essential step in determining where a function reaches its local maximum, minimum, or any turning points.

To identify critical points, we first calculate the derivative of a function and set it equal to zero to solve for the variable. This process reveals points where the function changes direction or stays flat.

• For example, with a function like \( y = \frac{x}{x^2 + 1} \), we find that its derivative is \( y' = \frac{1 - x^2}{(x^2 + 1)^2} \).
• Setting \( y' = 0 \) gives us the equation \( 1 - x^2 = 0 \), which simplifies to \( x^2 = 1 \).
• This solution leads us to our critical points at \( x = 1 \) and \( x = -1 \).

Critical points highlight possible candidate points for local extrema, but further testing, like the second derivative test, is needed to classify them fully.

Derivative

The derivative of a function represents the rate at which the function's value is changing at any given point. It’s a tool used to compute the slope of a function's graph, particularly the slope of the tangent line at any point.

To find the derivative, we often rely on rules such as the power rule, product rule, and especially the quotient rule when dealing with functions that are ratios of two functions. It tells us how to differentiate functions of the form \( \frac{u}{v} \), where \( u \) and \( v \) are functions of \( x \):
\[ \biggl( \frac{u}{v} \biggr)' = \frac{v \cdot u' - u \cdot v'}{v^2} \]
For the function \( y = \frac{x}{x^2 + 1} \), this rule was used to find the derivative:

• Let \( u = x \) and \( v = x^2 + 1 \)
• Then \( u' = 1 \) and \( v' = 2x \)
• The derivative \( y' = \frac{(x^2 + 1) \cdot 1 - x \cdot 2x}{(x^2 + 1)^2} = \frac{1 - x^2}{(x^2 + 1)^2} \)

Knowing the derivative allows us to explore the behavior of the function, especially near its critical points.

Second Derivative Test

The second derivative test is a method used to determine if a critical point is a local maximum, local minimum, or neither. By evaluating the second derivative of a function at its critical points, we get insights into the function's concavity, or how it curves.

If the second derivative \( y'' \) is:

• Positive at a critical point, the function is concave up, indicating a local minimum.
• Negative, the function is concave down, indicating a local maximum.
• If \( y'' \) is zero, the test is inconclusive, and other methods are required.

In the example \( y = \frac{x}{x^2 + 1} \), we calculated the second derivative to explore critical points further:

• At \( x = 1 \), \( y''(1) \) is negative, signifying a local maximum.
• At \( x = -1 \), \( y''(-1) \) is positive, indicating a local minimum.

The second derivative test helps classify and better understand local behaviors of the function at these points.

Quotient Rule

The quotient rule is a fundamental technique in calculus for differentiating functions that are ratios of two other functions. It's especially useful when each component of the fraction involves different algebraic expressions. This rule helps manage the complex derivatives and makes solving critical points feasible.

The quotient rule formula is:
\[\biggl( \frac{u}{v} \biggr)' = \frac{v \cdot u' - u \cdot v'}{v^2} \]
Where \( u \) and \( v \) are functions of \( x \), and \( u' \), \( v' \) are their respective derivatives. This rule applies seamlessly to the function \( y = \frac{x}{x^2 + 1} \).

• Identify \( u = x \) and \( v = x^2 + 1 \)
• Compute \( u' = 1 \) and \( v' = 2x \)
• Resulting in the derivative \( y' = \frac{1 - x^2}{(x^2 + 1)^2} \)

Through this approach, the quotient rule significantly simplifies the process of finding derivatives of complex fractions, enabling more efficient analysis of a function’s behavior.