Problem 15 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 5-26, identify the critical points and find the maximum value and minimum value on the given interval. $$ g(x)=\frac{1}{1+x^{2}} ; I=(-\infty, \infty) \text { Hint: Sketch the graph. } $$

Step-by-Step Solution

Answer

Max: 1 at \(x=0\); No minimum value.

Step 1: Find the Derivative

To find critical points, we first need the derivative of the function with respect to \(x\). For the given function \(g(x) = \frac{1}{1+x^2}\), use the quotient rule. The quotient rule states \(\left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2}\). Here, \(u = 1\) and \(v = 1+x^2\). So, \(u' = 0\) and \(v' = 2x\). Hence, the derivative is \(-\frac{2x}{(1+x^2)^2}\).

Step 2: Set Derivative to Zero

To find critical points, set the derivative \(-\frac{2x}{(1+x^2)^2}\) equal to zero. This simplifies to \(-2x = 0\), which implies \(x = 0\). Therefore, \(x = 0\) is a critical point.

Step 3: Test the Critical Points

Evaluate the function \(g(x) = \frac{1}{1+x^2}\) at the critical point \(x = 0\) and also examine the behavior as \(x\) approaches infinity and negative infinity. At \(x = 0\), \(g(0) = 1\). As \(x\) approaches \(\infty\) or \(-\infty\), \(g(x)\) approaches 0.

Step 4: Analyze the Graph

Since the hint suggests sketching the graph, note that \(g(x)\) is always positive, symmetric about \(x = 0\), and has a maximum at \(x = 0\). The graph has horizontal asymptotes at \(y=0\).

Step 5: Identify Maximum and Minimum Values

From the analysis, \(g(x)\) achieves a maximum value of 1 at \(x = 0\). There is no minimum value because \(g(x)\) tends asymptotically towards 0 as \(x\) approaches \(\infty\) or \(-\infty\). Therefore, the minimum on the interval is not achieved at a finite \(x\).

Key Concepts

Quotient Rule

The Quotient Rule is a valuable tool in calculus used to find the derivative of a function that is the ratio of two differentiable functions. In simpler terms, if you have a function divided by another, the Quotient Rule helps determine its derivative.
To apply the Quotient Rule, you need to identify these components: a numerator function \( u \) and a denominator function \( v \). The rule is expressed as:\[ \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \]In words, the derivative of the quotient is the derivative of the numerator multiplied by the denominator, minus the numerator multiplied by the derivative of the denominator, all divided by the square of the denominator.
For our function \( g(x) = \frac{1}{1 + x^2} \), applying the quotient rule:

• \( u = 1 \) and \( v = 1 + x^2 \).
• Calculating derivatives: \( u' = 0 \) and \( v' = 2x \).
• Substituting in the formula gives: \(-\frac{2x}{(1 + x^2)^2} \).

This derivative helps us further examine the critical points and behavior of the function later on.

Derivative Analysis

Derivative analysis involves examining the derivative of a function to understand aspects like critical points, increasing or decreasing behavior, and concavity. The critical points occur where the derivative is zero or undefined.
For the given function with derivative \(-\frac{2x}{(1 + x^2)^2} \), finding critical points involves setting the derivative equal to zero:\[ -2x = 0 \]This simplifies to \( x = 0 \).
With the critical point identified, we understand:

• At \( x = 0 \), the derivative is zero, indicating potential maxima or minima.
• Since the function's derivative does not have undefined points, no other critical points exist.

Further analysis of these points will help discern the function's behavior more accurately.

Function Behavior

Understanding the overall behavior of a function often combines insights from both derivatives and function evaluation. With \( g(x) = \frac{1}{1 + x^2} \), both critical points and end behavior matter.
Evaluate the function at critical points and limits:

• At \( x = 0 \), \( g(0) = 1 \), suggesting a peak or maximum value.
• As \( x \to \infty \) or \( x \to -\infty \), \( g(x) \to 0 \), indicating horizontal asymptotes at \( y = 0 \).

These evaluations show:

• The function is symmetric about the y-axis (even function).
• Peaks at \( x = 0 \) and approaches zero elsewhere, confirming the maximum and not minimum values.

By sketching the graph, you clearly see this behavior, with the function being positive and showing a smooth decline toward the asymptotes. Thus, the graph won't cross below the x-axis, maintaining a defined region for maximum and approaching behaviors.