Problem 18 - Calculus Early Transcendentals

Question

Find: (a) the intervals on which f is increasing, (b) the intervals on which f is decreasing, (c) the open intervals on which f is concave up, (d) the open intervals on which f is concave down, and (e) the x-coordinates of all inflection points. \(f(x)=5+12 x-x^{3}\)

Step-by-Step Solution

Answer

f is increasing on (-2, 2); decreasing on (-∞, -2) and (2, ∞); concave up on (-∞, 0); concave down on (0, ∞); inflection point at (0, 5).

Step 1: Find the derivative

First, we need to find the first derivative of the function to determine where it is increasing or decreasing. Given function: \[ f(x) = 5 + 12x - x^3 \]First derivative:\[ f'(x) = \frac{d}{dx}(5 + 12x - x^3) = 12 - 3x^2 \]

Step 2: Critical points and intervals of increase/decrease

Find the critical points by setting the first derivative equal to zero and solving for \(x\).\[ 12 - 3x^2 = 0 \]Solve for \(x\):\[ 3x^2 = 12 \]\[ x^2 = 4 \] \[ x = \pm 2 \]Test intervals around the critical points \(x = -2\) and \(x = 2\) to determine where \(f(x)\) is increasing or decreasing:- For \(x < -2\), pick \(x = -3\), \(f'(-3) = 12 - 3(-3)^2 = 12 - 27 = -15\) (decreasing)- For \(-2 < x < 2\), pick \(x = 0\), \(f'(0) = 12\) (increasing)- For \(x > 2\), pick \(x = 3\), \(f'(3) = 12 - 27 = -15\) (decreasing)Thus, \(f(x)\) is increasing on \((-2, 2)\) and decreasing on \((-\infty, -2)\) and \((2, \infty)\).

Step 3: Find the second derivative

Next, find the second derivative to determine the concavity of \(f(x)\).Second derivative:\[ f''(x) = \frac{d}{dx}(12 - 3x^2) = -6x \]

Step 4: Test for concavity

Set the second derivative equal to zero to find possible inflection points:\[ -6x = 0 \]\[ x = 0 \]Test intervals around \(x = 0\) to determine concavity:- For \(x < 0\), pick \(x = -1\), \(f''(-1) = -6(-1) = 6\) (concave up)- For \(x > 0\), pick \(x = 1\), \(f''(1) = -6(1) = -6\) (concave down)Thus, \(f(x)\) is concave up on \((-\infty, 0)\) and concave down on \((0, \infty)\).

Step 5: Determine inflection points

An inflection point occurs where the concavity changes. Since \(x = 0\) is where the concavity of \(f(x)\) changes from concave up to concave down, it is an inflection point. Evaluate \(f(x)\) at \(x = 0\):\[ f(0) = 5 + 12(0) - 0^3 = 5 \]Thus, the inflection point is at \((0, 5)\).

Key Concepts

Derivative Analysis

The first step in tackling any calculus problem involving increasing or decreasing functions is derivative analysis. This involves finding the derivative of the given function. The derivative, denoted as \( f'(x) \), gives us critical information about the slope of the function at any point. For the function \( f(x) = 5 + 12x - x^3 \), we calculate the first derivative to explore these changes:

• Function: \( f(x) = 5 + 12x - x^3 \)
• First Derivative: \( f'(x) = 12 - 3x^2 \)

By understanding what the derivative tells us, students can determine whether the function \( f(x) \) is increasing, decreasing, or remaining constant over particular intervals. This forms the foundation needed to move on to analyzing other features of the graph, such as critical points and concavity.

Critical Points

Critical points are key values of \( x \) where the derivative \( f'(x) \) equals zero or is undefined. At these points, the function's behavior changes, indicating potential maximums, minimums, or points of inflection. By setting \( f'(x) = 12 - 3x^2 \) to zero, we find:

• Critical points occur at \( x = \pm 2 \)

After identifying the critical points, students can also determine on which intervals the function is increasing or decreasing:- Test intervals around \( x = -2 \) and \( x = 2 \): - \( f'(x) < 0 \) when \( x < -2 \) and \( x > 2 \) (decreasing intervals). - \( f'(x) > 0 \) between \( x = -2 \) and \( x = 2 \) (increasing interval).Understanding critical points helps students predict the behavior of the function across its domain and analyze the nature of these points.

Concavity

Concavity involves understanding the curvature of a function, which is determined via the second derivative. The second derivative, denoted \( f''(x) \), helps determine whether a function is concave up or concave down over certain intervals. For \( f(x) = 5 + 12x - x^3 \), the second derivative is:

• Second derivative: \( f''(x) = -6x \)

By assessing the sign of \( f''(x) \), we determine concavity:- If \( f''(x) > 0 \), the curve is concave up.- If \( f''(x) < 0 \), the curve is concave down.Testing around potential inflection points, we understand:- The function is concave up when \( x < 0 \) and concave down when \( x > 0 \).These insights allow students to visualize how the function curves across different values of \( x \) and inform the overall shape of the graph.

Inflection Points

An inflection point is where the function changes its concavity, switching from concave up to concave down, or vice versa. This change is particularly noticeable where \( x \)-values make the second derivative go from positive to negative or vice versa. For the function \( f(x) = 5 + 12x - x^3 \), solve \( f''(x) = -6x = 0 \) to find:

• Inflection point: \( x = 0 \)

At this point, concavity shifts:- From concave up on \( (-\infty, 0) \) to concave down on \( (0, \infty) \).Inflection points provide critical insight into graphing the function, ensuring students understand where curves shift direction. Evaluating \( f(x) \) at the inflection point \( x = 0 \), we find its coordinate is \( (0, 5) \). This leads to a fuller analysis and presentation of the function's key features.