Problem 33 - Calculus Early Transcendentals: Pearson New International Edition

Question

Sketch the graph of a continuous function fon \([0,6]\) that satisfies all the stated conditions. $$ \begin{array}{l} f(0)=f(4)=1 ; f(2)=2 ; f(6)=0 ; \\ f^{\prime}(x)>0 \text { on }(0,2) ; f^{\prime}(x)<0 \text { on }(2,4) \cup(4,6) ; \\ f^{\prime}(2)=f^{\prime}(4)=0 ; f^{\prime \prime}(x)>0 \text { on }(0,1) \cup(3,4) \\ f^{\prime \prime}(x)<0 \text { on }(1,3) \cup(4,6) \end{array} $$

Step-by-Step Solution

Answer

Sketch a curve passing through the points (0,1), (2,2), (4,1), and (6,0) that increases from 0 to 2, then decreases to 4 and further to 6, with specified concavity.

Step 1: Plot Given Points

Identify and plot the points where the function has specific values, according to the conditions. The points are \((0, 1), (2, 2), (4, 1), (6, 0)\). These points will lie on the curve of the function \(f(x)\).

Step 2: Identify Increasing and Decreasing Intervals using First Derivative

From the information given, \(f'(x) > 0\) on \((0,2)\), meaning the function is increasing in this interval. Similarly, \(f'(x) < 0\) on \((2,4)\) and \((4,6)\), indicating the function is decreasing in these intervals. This will help us understand the shape of the graph between and after the plotted points.

Step 3: Analyzing Critical Points with First Derivative

The critical points occur where \(f'(x) = 0\). These are given at \(x = 2\) and \(x = 4\). These are the points where the graph changes its direction or has a horizontal tangent.

Step 4: Determine Concavity using Second Derivative

Analyze the intervals given for the second derivative: \(f''(x) > 0\) on \((0,1)\) and \((3,4)\), indicating that the graph is concave up in these intervals (looks like a U). \(f''(x) < 0\) on \((1,3)\) and \((4,6)\), indicating that the graph is concave down in these intervals (looks like an upside-down U).

Step 5: Sketch the Graph

Begin sketching the graph, following these steps:1. Draw an increasing curve from \((0, 1)\) to \((2, 2)\), concave up from \(x=0\) to \(x=1\) and concave down from \(x=1\) to \(x=2\).2. At \(x=2\), the tangent is horizontal, marking a local maximum.3. Draw a decreasing curve from \((2, 2)\) to \((4, 1)\), concave down from \(x=2\) to \(x=4\), with another horizontal tangent at \(x=4\).4. Draw a decreasing curve from \((4, 1)\) to \((6, 0)\), concave down. Ensure continuity between all sections.

Key Concepts

First Derivative Test

The First Derivative Test is a method used to determine where a function is increasing or decreasing. By evaluating the first derivative of a function, denoted as \( f'(x) \), we can gather important information about its behavior. If \( f'(x) > 0 \) across an interval, the function is increasing on that interval. Conversely, if \( f'(x) < 0 \), the function is decreasing. These insights are crucial when sketching the graph of a function.

In the original exercise, we analyze \( f'(x) \) to find intervals of increase and decrease:

• On \((0,2)\), \( f'(x) > 0 \), indicating that the graph of the function rises.
• On \((2,4)\) and \((4,6)\), \( f'(x) < 0 \), so the function is decreasing in these intervals.

Understanding the behavior of the function through its derivative helps us to sketch more accurate graphs and visualize how the function behaves in different intervals.

Second Derivative Test

The Second Derivative Test provides insights into the concavity of a function and helps identify points of inflection, where the concavity changes. The second derivative, \( f''(x) \), tells us whether a function is curving upwards or downwards:

• If \( f''(x) > 0 \), the graph is concave up, creating a U-shape.
• If \( f''(x) < 0 \), the graph is concave down, resembling an upside-down U.

In the exercise, concavity is analyzed as follows:

• \( f''(x) > 0 \) on \((0,1)\) and \((3,4)\), meaning the graph is concave up in these intervals.
• \( f''(x) < 0 \) on \((1,3)\) and \((4,6)\), indicating the graph is concave down.

Recognizing the concavity allows us to predict how the graph bends and helps us capture the right curvature when sketching lines on a plot.

Critical Points

Critical points are points on the graph where the derivative of the function is zero or undefined. These points are significant because they can often indicate local maxima, minima, or saddle points of the function.

In the problem given, we find the critical points where \( f'(x) = 0 \), specifically at \( x = 2 \) and \( x = 4 \). These critical points represent potential turning points where the function's slope reverses, changing from increasing to decreasing, or vice versa. In our specific exercise:

• At \( x = 2 \), the function has a horizontal tangent, suggesting a local maximum since it transitions from increasing to decreasing.
• At \( x = 4 \), another horizontal tangent occurs, marking a point where the function stops decreasing before continuing to decrease, typically highlighting an inflection or steady point.

Recognizing and analyzing critical points are vital steps in accurately sketching the function's overall shape.

Concavity and Inflections

Concavity refers to the direction in which a graph curves, while points of inflection are where the concavity changes. Understanding these characteristics is essential as they affect how we interpret the graph's shape.

In the exercise, the intervals of concavity are identified using the second derivative:

• Concave up on \((0,1)\) and \((3,4)\): This means the graph forms a U-shape, curving upwards.
• Concave down on \((1,3)\) and \((4,6)\): Here, the graph curves downwards, taking on an inverted U-shape.

Points of inflection occur where the graph transitions from one type of concavity to another. These transitions are critical for properly sketching and understanding the dynamics of the graph.
By applying this knowledge, one can predict how the graph will look, allowing for a more accurate and intuitive sketch of the function within the given domain [0,6].