Problem 19 - 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). \(f(x)=x^{3}-12 x+1\)

Step-by-Step Solution

Answer

The function increases on (-∞, -2) and (2, ∞), decreases on (-2, 2), is concave down on (-∞, 0), and concave up on (0, ∞).

Step 1: Find the Critical Points

First, find the first derivative of the function.\[ f'(x) = \frac{d}{dx}(x^3 - 12x + 1) = 3x^2 - 12. \]Set the first derivative equal to zero to find the critical points:\[ 3x^2 - 12 = 0. \]This simplifies to:\[ x^2 = 4, \]so the critical points are \(x = 2\) and \(x = -2\).

Step 2: Determine Intervals of Increase and Decrease

Evaluate the first derivative at points in the intervals determined by the critical points (-∞, -2), (-2, 2), and (2, ∞).- For \(x < -2\) (e.g., \(x = -3\)): \[ f'(-3) = 3(-3)^2 - 12 = 27 - 12 = 15 > 0, \] so the function is increasing.- For \(-2 < x < 2\) (e.g., \(x = 0\)): \[ f'(0) = 3(0)^2 - 12 = -12 < 0, \] so the function is decreasing.- For \(x > 2\) (e.g., \(x = 3\)): \[ f'(3) = 3(3)^2 - 12 = 27 - 12 = 15 > 0, \] so the function is increasing.

Step 3: Determine Concavity and Points of Inflection

Find the second derivative:\[ f''(x) = \frac{d}{dx}(3x^2 - 12) = 6x. \]Set the second derivative equal to zero to find potential inflection points:\[ 6x = 0, \]so \(x = 0\) is a potential inflection point.Evaluate the second derivative at points in the intervals (-∞, 0) and (0, ∞):- For \(x < 0\): \[ f''(-1) = 6(-1) = -6 < 0, \] so the function is concave down.- For \(x > 0\): \[ f''(1) = 6(1) = 6 > 0, \] so the function is concave up.

Step 4: Sketch the Graph

Using the information from previous steps, sketch the graph of \(f(x) = x^3 - 12x + 1\) indicating:- The function increases on \((-∞, -2)\) and \((2, ∞)\).- The function decreases on \((-2, 2)\).- The function is concave down on \((-∞, 0)\) and concave up on \((0, ∞)\).- There is an inflection point at \(x = 0\), and critical points at \(x = -2\) and \(x = 2\).

Key Concepts

Critical Points

In calculus, critical points are the values of \( x \) where a function's derivative is zero or undefined. These points are significant because they often signal where the graph of a function changes direction, such as where a peak or a trough might occur. For the function \( f(x) = x^3 - 12x + 1 \), critical points occur where the first derivative, \( f'(x) = 3x^2 - 12 \), is equal to zero. Solving

• \( 3x^2 - 12 = 0 \)
• gives \( x = 2 \) and \( x = -2 \).

These values are our critical points, where the behavior of the graph might shift from increasing to decreasing or vice versa. Knowing critical points helps to piece together the overall behavior of the graph.

Increasing and Decreasing Intervals

Understanding where a function is increasing or decreasing is fundamental in analyzing its behavior. To determine this, we look at the sign of the first derivative within each interval defined by the critical points.

• For \( x < -2 \), the derivative \( f'(-3) = 15 \), meaning the function is increasing.
• Between \( x = -2 \) and \( x = 2 \), \( f'(0) = -12 \), indicating the function is decreasing.
• For \( x > 2 \), \( f'(3) = 15 \), showing the function is increasing again.

Intervals where the function increases indicate an upward slope, while decreases describe a downward slope. This helps in sketching the graph and understanding the overall trend of the curve.

Concavity and Points of Inflection

Concavity tells you about the curvature of the graph — whether it curves upwards or downwards. To find this, we use the second derivative. For \( f(x) = x^3 - 12x + 1 \), the second derivative is \( f''(x) = 6x \).

• If \( f''(x) > 0 \), the graph is concave up (like a smile).
• If \( f''(x) < 0 \), the graph is concave down (like a frown).

At \( x = 0 \), \( f''(x) = 0 \), which is a potential point of inflection where the concavity changes. Further evaluation confirms:

• For \( x < 0 \), \( f''(-1) = -6 \). The graph is concave down.
• For \( x > 0 \), \( f''(1) = 6 \). The graph is concave up.

Thus, \( x = 0 \) is a point of inflection, where the curve changes its concavity, giving us further insight into the nature of the function's graph.

Derivative Analysis

Derivative analysis involves examining both the first and second derivatives of a function to gather information about the function's graph. By analyzing derivatives, we can determine critical points, increasing/decreasing intervals, and concavity.

• The first derivative, \( f'(x) = 3x^2 - 12 \), tells where the graph is rising or falling.
• The second derivative, \( f''(x) = 6x \), provides insights into the graph’s curvature.

This dual approach gives a comprehensive picture of the function’s behavior. Derivative analysis is crucial in calculus, as it lays the groundwork for deeper exploration and understanding of how functions behave.