Problem 24 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 19-28, determine where the graph of the given function is increasing, decreasing, concave up, and concave down. Then sketch the graph (see Example 4). \(H(x)=\frac{x^{2}}{x^{2}+1}\)

Step-by-Step Solution

Answer

The function is increasing on \((0, \infty)\), decreasing on \((-\infty, 0)\), concave up on \((-\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}})\), and concave down elsewhere.

Step 1: Find the First Derivative

To determine where the function is increasing or decreasing, find the first derivative of \(H(x)\). For \(H(x)=\frac{x^{2}}{x^{2}+1}\), apply the quotient rule: \[ H'(x) = \frac{(x^{2}+1)(2x) - (x^{2})(2x)}{(x^{2}+1)^2} = \frac{2x}{(x^2+1)^2}.\] The simplified form of the derivative is \(H'(x) = \frac{2x}{(x^2+1)^2}\).

Step 2: Determine Where the Function is Increasing or Decreasing

The function is increasing where \(H'(x) > 0\) and decreasing where \(H'(x) < 0\). Since the denominator \((x^2+1)^2\) is always positive, consider the numerator:\(2x > 0\), which implies \(x > 0\). Thus, \(H(x)\) is increasing on \((0, \infty)\) and decreasing on \((-\infty, 0)\).

Step 3: Find the Second Derivative

To find concavity, calculate the second derivative of \(H(x)\). Start with \(H'(x) = \frac{2x}{(x^2+1)^2}\) and use the quotient rule again: \[ H''(x) = \frac{((x^2+1)^2)(2) - (2x)(4x(x^2+1))}{(x^2+1)^4} = \frac{2(x^2+1)^2 - 8x^2(x^2+1)}{(x^2+1)^4}.\] Simplify to get \(H''(x) = \frac{2(x^2+1 - 4x^2)}{(x^2+1)^3} = \frac{2(1-3x^2)}{(x^2+1)^3}.\)

Step 4: Determine Concavity

Analyze \(H''(x) = \frac{2(1-3x^2)}{(x^2+1)^3}\). The numerator \(2(1-3x^2)\) determines the concavity:- \(H(x)\) is concave up where \(H''(x) > 0\), i.e., \(1-3x^2 > 0\), or \(-\frac{1}{\sqrt{3}} < x < \frac{1}{\sqrt{3}}\).- \(H(x)\) is concave down where \(H''(x) < 0\), i.e., \(x < -\frac{1}{\sqrt{3}}\) or \(x > \frac{1}{\sqrt{3}}\).

Key Concepts

First Derivative

The first derivative of a function tells us about the rate of change at any given point. It's like looking at how fast or slow a car is going without worrying about where it's headed. For the function \(H(x) = \frac{x^{2}}{x^{2}+1}\), we calculate the first derivative using the quotient rule. This rule is applied when you have two functions divided by each other, like a "top" function and a "bottom" function. To find \(H'(x)\), we used the formula for the derivative of a quotient:

• Derivative of the top \((x^2)\) is \(2x\).
• Derivative of the bottom \((x^2+1)\) is \(2x\).

After applying the quotient rule, we simplify the result to \(H'(x) = \frac{2x}{(x^2+1)^2}\). The first derivative helps determine whether the function is increasing or decreasing at different intervals.

Function Increasing and Decreasing

Once we have the first derivative \(H'(x) = \frac{2x}{(x^2+1)^2}\), we can find out where the function is increasing or decreasing. Think of increasing as going uphill and decreasing as going downhill. To find these intervals:

• The function is increasing where \(H'(x) > 0\).
• The function is decreasing where \(H'(x) < 0\).

Since the denominator \((x^2+1)^2\) is always positive, we just solve \(2x > 0\): - It means \(x > 0\), so the function is increasing on \((0, \infty)\).- If \(x < 0\), then the function is decreasing on \((-\infty, 0)\).These insights into intervals help sketch the behavior of the graph over a series of points.

Second Derivative

The second derivative tells us about the concavity of the graph. It is sort of like understanding whether the path is a bowl shape or a cap. To begin, we find the second derivative of \(H(x)\) from the first derivative \(H'(x) = \frac{2x}{(x^2+1)^2}\).Applying the quotient rule again, we get:\[ H''(x) = \frac{2(x^2+1)^2 - 8x^2(x^2+1)}{(x^2+1)^4} \]Simplify to get:\[ H''(x) = \frac{2(1-3x^2)}{(x^2+1)^3} \]Finding this derivative helps us determine how the graph curves at different points, and this is essential for understanding the nature of graph fluctuations.

Concavity

Concavity describes whether the function is curving upwards or downwards. When analyzing \(H''(x) = \frac{2(1-3x^2)}{(x^2+1)^3}\), we focus on the numerator, \(2(1-3x^2)\), to find where the function is concave up or concave down.- **Concave Up:** The graph is concave up where \(H''(x) > 0\). This occurs when \(1-3x^2 > 0\), or for \(-\frac{1}{\sqrt{3}} < x < \frac{1}{\sqrt{3}}\).- **Concave Down:** The graph is concave down where \(H''(x) < 0\). This situation arises for \(x < -\frac{1}{\sqrt{3}}\) or \(x > \frac{1}{\sqrt{3}}\).Understanding concavity helps us anticipate the direction in which a function will curve, informing how we predict its behavior over time. Such details are crucial when visualizing complex curves and predicting maximum or minimum points.