Problem 24 - Calculus Early Transcendentals: Pearson New International Edition

Question

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 decreasing on \((-\infty, 0)\), increasing on \((0, \infty)\), and concave up near 0; sketch accordingly.

Step 1: Find the First Derivative

To determine where the function is increasing or decreasing, we first need its first derivative, denoted as \( H'(x) \). For the function \( H(x)=\frac{x^{2}}{x^{2}+1} \), use the quotient rule: \( u = x^2 \), \( v = x^2 + 1 \), hence \( u' = 2x \) and \( v' = 2x \). The quotient rule gives us:\[H'(x) = \frac{(v)(u') - (u)(v')}{v^2} = \frac{(x^2 + 1)(2x) - (x^2)(2x)}{(x^2 + 1)^2} = \frac{2x}{(x^2 + 1)^2}\]

Step 2: Find Critical Points

Set the first derivative equal to zero to find critical points:\[2x = 0\]Solving for \(x\), we find \(x = 0\). This is our critical point, which we will use to determine intervals of increase and decrease.

Step 3: Test Intervals for Increasing/Decreasing

Choose test points in the intervals created by the critical point to determine whether the function is increasing or decreasing. Test \(x = -1\) and \(x = 1\):- For \(x = -1\), \(H'(-1) = \frac{2(-1)}{((-1)^2 + 1)^2} = \frac{-2}{4} = -0.5\), implying decreasing.- For \(x = 1\), \(H'(1) = \frac{2(1)}{(1^2 + 1)^2} = \frac{2}{4} = 0.5\), implying increasing.Thus, \(H(x)\) is decreasing on \((-\infty, 0)\) and increasing on \((0, \infty)\).

Step 4: Find the Second Derivative

To determine concavity, find the second derivative. Differentiate \(H'(x)\):\[H''(x)\] can be calculated using the quotient rule again:\(u = 2x\), \(v = (x^2 + 1)^2\), so \(u' = 2\) and \(v' = 4x(x^2 + 1)\):\[H''(x) = \frac{(v)(u') - (u)(v')}{v^2} = \frac{((x^2+1)^2)(2) - (2x)(4x(x^2+1))}{(x^2+1)^4}\]Simplifying:\[H''(x) = \frac{2(x^2+1)^2 - 8x^2(x^2+1)}{(x^2+1)^4}\]

Step 5: Simplify the Second Derivative

Continue simplifying the second derivative:\[= \frac{2(x^4 + 2x^2 + 1) - 8x^2(x^2 + 1)}{(x^2+1)^4} = \frac{2x^4 + 4x^2 + 2 - 8x^4 - 8x^2}{(x^2+1)^4}\]\[H''(x) = \frac{-6x^4 - 4x^2 + 2}{(x^2+1)^4}\]

Step 6: Determine Concavity

Set \( H''(x) = 0 \) or analyze the sign of \( H''(x) \).For simplification, determine if the nominator can change sign:1. Find the roots of \(-6x^4 - 4x^2 + 2 = 0\) or check values: - For \(x = 0\), \(H''(0) = \frac{2}{1} > 0\), indicating concave up. - For \(x^2 > 1\), check a value such as 2, leads to \(-6 \times 16 - 4 \times 4 + 2\), which should be < 0, indicating concave down. Final conclusion: concave up in a closer vicinity around 0.

Step 7: Sketch the Graph

Using the intervals: - The function is decreasing on \((-\infty, 0)\) and increasing on \((0, \infty)\).- Concave up near 0, hence increases and decreases smoothly around this point.- Expect concave down starts as \(x\) and \(-x\) move beyond 0.Plot these behaviors to intuitively sketch the function.

Key Concepts

Increasing and Decreasing Functions

To understand when a function is increasing or decreasing, we rely on its first derivative. When the first derivative, denoted as \(f'(x)\), is positive at a point, the function is increasing at that interval. Likewise, if \(f'(x)\) is negative, the function is decreasing.

For our example function, \(H(x) = \frac{x^2}{x^2 + 1}\), we found the first derivative using the quotient rule:

• \(H'(x) = \frac{2x}{(x^2 + 1)^2}\)

By setting \(H'(x) = 0\), we identified critical points where the function could change from increasing to decreasing or vice versa. In this case, \(x = 0\) is a critical point.

We then tested intervals around this point, choosing values such as \(x = -1\) and \(x = 1\):

• At \(x = -1\), \(H'(-1)\) was negative, hence the function is decreasing at \((-\infty, 0)\).
• At \(x = 1\), \(H'(1)\) was positive, thus the function increases at \((0, \infty)\).

Thus, \(H(x)\) is decreasing on \((-\infty, 0)\) and increasing on \((0, \infty)\).

Concavity of Functions

Concavity describes the curvature of a graph. A function is concave up when it "holds water" like a cup, and concave down when it "spills water".

To identify concavity, we examine the second derivative, \(f''(x)\). If \(f''(x) > 0\), the graph is concave up. If \(f''(x) < 0\), it is concave down.

For the function \(H(x)\), we calculated its second derivative:

• \(H''(x) = \frac{-6x^4 - 4x^2 + 2}{(x^2+1)^4}\)

Evaluating \(H''(x)\) helps determine which intervals are concave up or down.

We checked values for \(x = 0\) and found \(H''(0) > 0\), signifying the function is concave up around \(x = 0\). Conversely, for \(x^2 > 1\), \(H''(x)\) becomes negative, indicating concavity shifts to downwards.

First Derivative Test

The first derivative test helps us determine where a function is increasing or decreasing and identifying local maxima and minima.

We look at the sign changes in the first derivative \(f'(x)\) at a critical point. A change from positive to negative indicates a local maximum, whereas a change from negative to positive suggests a local minimum.

In the case of our function \(H(x)\), we previously saw that \(H'(x)\) changes sign at \(x = 0\):

• From negative to positive, indicating a local minimum at \(x = 0\).

Thus, the first derivative test confirms that \(x = 0\) is where \(H(x)\) switches from decreasing to increasing.

This allows us to predict not only growth behavior of the function but also its local optimization tendencies in a plot.

Second Derivative Test

The second derivative test complements the first derivative test by giving a deeper insight into concavity and identifying critical points more skillfully when \(f'(x) = 0\).

By examining the second derivative \(f''(x)\) at critical points:

• If \(f''(x) > 0\), the point is a local minimum (concave up).
• If \(f''(x) < 0\), the point is a local maximum (concave down).
• If \(f''(x) = 0\), the test is inconclusive.

For our function \(H(x)\), at the critical point \(x = 0\), \(H''(0) > 0\), confirming a local minimum. This verification supports the conclusion from the first derivative test.

Thus, the second derivative test helps affirm our earlier findings and more accurately maps the behavior of the function, especially its bends and turns.