Problem 34 - Essential Calculus Early Transcendentals

Question

(a) Find the intervals of increase or decrease. (b) Find the local maximum and minimum values. (c) Find the intervals of concavity and the inflection points. (d) Use the inforvation from parts (a)-(c) to sketch the graph. Check your work with a graphing device if you have one. \(f(x)=\ln \left(x^{4}+27\right)\)

Step-by-Step Solution

Answer

Increase on \((0, \infty)\); decrease on \((-\infty, 0)\); local min at \(x=0\); concavity changes suggest inflection at \(x\approx\pm6.36\).

Step 1: Find the First Derivative

To find the intervals of increase or decrease, start by finding the derivative of the function, \( f(x) \). The function is \( f(x) = \ln(x^4 + 27) \). Using the chain rule, the derivative is \( f'(x) = \frac{1}{x^4 + 27} \times 4x^3 = \frac{4x^3}{x^4 + 27} \).

Step 2: Determine Intervals of Increase or Decrease

Set the first derivative equal to zero to find critical points: \( \frac{4x^3}{x^4 + 27} = 0 \Rightarrow x = 0 \). Test the intervals \((-\infty, 0)\) and \((0, \infty)\) using test points in the first derivative. If \( x < 0 \), \( f'(x) < 0 \), and if \( x > 0 \), \( f'(x) > 0 \). The function decreases on \((-\infty, 0)\) and increases on \((0, \infty)\).

Step 3: Find the Second Derivative

Next, find the second derivative to determine concavity and inflection points. Start by deriving \( f'(x) \):\[ f''(x) = \frac{d}{dx} \left( \frac{4x^3}{x^4 + 27} \right) \]. Use the quotient rule, \( v = x^4 + 27 \) and \( u = 4x^3 \), to find \[ f''(x) = \frac{(x^4 + 27)(12x^2) - (4x^3)(4x^3)}{(x^4 + 27)^2} = \frac{12x^6 + 324x^2 - 16x^6}{(x^4 + 27)^2} = \frac{-4x^6 + 324x^2}{(x^4 + 27)^2} \].

Step 4: Determine Intervals of Concavity

Set the second derivative equal to zero to find potential inflection points: \( -4x^6 + 324x^2 = 0 \Rightarrow x^2(4x^4 - 324) = 0 \). Solve \( x^2 = 0 \) giving \( x = 0 \), and \( 4x^4 = 324 \) gives \( x = \pm \sqrt{\frac{81}{2}} \approx \pm 6.36 \). Using sign analysis, determine that the function is concave up on \((-\infty, -6.36)\) and \((0, 6.36)\), and concave down on \((-6.36, 0)\) and \((6.36, \infty)\).

Step 5: Identify Local Maximum and Minimum Values

Using the results from Steps 2 and 4, determine that there is a local minimum at \( x = 0 \), because the function changes from decreasing to increasing. There are no local maxima in the function.

Step 6: Use Information to Sketch the Graph

Use the intervals of increase/decrease, concavity, and critical points to sketch the graph. The graph decreases for \( x < 0 \) and increases for \( x > 0 \). The graph is concave up when \( x < -6.36 \) and \( 0 < x < 6.36 \), and it is concave down on \(-6.36 < x < 0 \) and \( x > 6.36 \). Check the graph manually or using a graphing device for verification.

Key Concepts

Intervals of Increase or Decrease

In calculus, understanding intervals of increase or decrease helps determine where a function is rising or falling. This begins with finding the first derivative of the function. For the given function, \( f(x) = \ln(x^4 + 27) \), we calculated the first derivative as \( f'(x) = \frac{4x^3}{x^4 + 27} \). To find where the function increases or decreases, we set \( f'(x) = 0 \), which gives the critical point \( x = 0 \). By testing intervals around this critical point, we establish that the function decreases for \( x < 0 \) since \( f'(x) < 0 \) in this range, and increases for \( x > 0 \) because \( f'(x) > 0 \) in this interval. These findings allow us to conclude that our function has one interval of decrease, \((-\infty, 0)\), and one of increase, \((0, \infty)\). Recognizing these patterns is crucial for analyzing the behavior of functions over their domains.

Critical Points

Critical points are essential in determining the locations of local maxima or minima of a function. By analyzing the first derivative, we identify points where the function's slope is zero, indicating potential peaks or troughs. For our function, \( f'(x) = \frac{4x^3}{x^4 + 27} = 0 \) at \( x = 0 \). Once we find these critical points, we must decide whether they correspond to a minimum, maximum, or neither. Here, since the function transitions from decreasing when \( x < 0 \) to increasing when \( x > 0 \), we determine there is a local minimum at \( x = 0 \). This insight is particularly useful when sketching graphs or solving optimization problems, as it guides us to the most significant points of interest on the graph.

Concavity and Inflection Points

Concavity describes how the slope of a function changes, either curving upwards or downwards, and is determined by the second derivative. We found the second derivative of our function as \( f''(x) = \frac{-4x^6 + 324x^2}{(x^4 + 27)^2} \). Solving \( f''(x) = 0 \), we identify potential inflection points at \( x = 0 \) and \( x \approx \pm 6.36 \).Conducting sign analysis on the second derivative, we establish the function's concavity:

• The function is concave up on \( (-\infty, -6.36) \) and \( (0, 6.36) \).
• It is concave down on \( (-6.36, 0) \) and \( (6.36, \infty) \).

An inflection point is where the function changes concavity. Thus, our function changes concavity at \( x = -6.36, 0, \) and \( 6.36 \), marking these as inflection points. Understanding these characteristics helps in accurately sketching and analyzing the graph of the function.

Graph Sketching

Graph sketching enables visualization of how the function behaves across its domain, incorporating information from critical points, intervals of increase/decrease, and concavity changes. Starting with known intervals: the function decreases before \( x = 0 \) and increases afterward.Next, inflection points indicate where the graph switches from curving up to down, informing us the function's key bends at \( x \approx \pm 6.36 \) and \( x = 0 \). This reflects:

• Concavity up for \((-\infty, -6.36)\)
• Concavity down for \((-6.36, 0)\)
• Concavity up for \((0, 6.36)\)
• And finally, concavity down for \((6.36, \infty)\)

For clarity and precision, sketching the graph by incorporating these intervals ensures a complete visualization. Remember, verifying your sketch with a graphing device can confirm accuracy, allowing for a more robust understanding of the function's behavior.