Problem 34 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 29-34, sketch the graph of a continuous function fon \([0,6]\) that satisfies all the stated conditions. \(f(0)=f(3)=3 ; f(2)=4 ; f(4)=2 ; f(6)=0\); $$ \begin{aligned} &f^{\prime}(x)>0 \text { on }(0,2) ; f^{\prime}(x)<0 \text { on }(2,4) \cup(4,5) \\ &f^{\prime}(2)=f^{\prime}(4)=0 ; f^{\prime}(x)=-1 \text { on }(5,6) \\ &f^{\prime \prime}(x)<0 \text { on }(0,3) \cup(4,5) ; f^{\prime \prime}(x)>0 \text { on }(3,4) \end{aligned} $$

Step-by-Step Solution

Answer

Graph is increasing to \(x=2\), decreasing after \(x=2\), and has specific critical points at \(x=2\) and \(x=4\).

Step 1: Identify Key Points and Values

Begin by marking the key points on the x-axis based on the given conditions: \(f(0) = 3\), \(f(2) = 4\), \(f(3) = 3\), \(f(4) = 2\), and \(f(6) = 0\). These points will help us shape the overall graph of the function \(f(x)\).

Step 2: Analyze Increasing and Decreasing Intervals

The derivative \(f'(x)\) controls the increase or decrease of the function. According to the conditions: \(f'(x) > 0\) on \((0,2)\), resulting in an increasing function. \(f'(x) < 0\) on \((2,4) \cup (4,5)\), which means the function decreases on these intervals. Additionally, \(f'(5,6) = -1\), which implies a line with a constant slope of -1 on this interval.

Step 3: Determine Critical Points

Critical points occur where \(f'(x)=0\). Based on the given conditions, \(f'(2)=0\) and \(f'(4)=0\). This suggests that the function has a local maximum at \(x=2\) since \(f'(x)\) changes signs from positive to negative, and a potential inflection point at \(x=4\) because \(f'(x)\) changes signs again.

Step 4: Examine Concavity

Examine the second derivative \(f''(x)\) to assess concavity. Given that \(f''(x) < 0\) on \((0,3) \cup (4,5)\), the graph is concave down in these ranges. With \(f''(x) > 0\) on \((3,4)\), the graph is concave up in this range. Therefore, the point at \(x=3\) is an inflection point as the concavity changes.

Step 5: Sketch the Graph

Using all the information: start at \(f(0)=3\), increase to \(f(2)=4\), decrease to \(f(3)=3\) with an inflection point at \(x=3\), continue decreasing to \(f(4)=2\) with a local minimum at \(x=4\), and further decrease to \(f(5)\) according to the slope \(f'(x)=-1\) which continues to \(f(6)=0\). Be sure to incorporate the concavity changes in your sketch.

Key Concepts

Continuous Function

A continuous function is a type of function that is unbroken or smooth. For any given interval, you can draw the graph of a continuous function without lifting your pencil from the paper. In math terms, this means that the function has no jumps, gaps, or abrupt changes in direction.
Understanding this concept is important because continuous functions behave predictably. They have no interruptions, which makes them very reliable for analysis. For instance, in the exercise we are discussing, the function is continuous on the interval \([0, 6]\).
This means that at every point between 0 and 6, the function is smoothly connected. You can easily track how the function transitions from one value to another without any sudden leaps or falls in the graph.

Derivative

The derivative of a function is a principal concept in calculus that measures how a function changes as its input changes. Simply put, it indicates the slope of the tangent line to the graph of the function at any particular point.
In the context of the exercise, derivative symbols like \(f'(x)\) help in understanding where the graph of the function is increasing or decreasing.

• If \(f'(x) > 0\), it means the function is increasing at that interval.
• If \(f'(x) < 0\), then the function is decreasing.

For example, in our exercise, the function increases on \((0,2)\) and decreases on \((2,4)\) and \((4,5)\).
The derivative also indicates constant slopes. When \(f'(x)\) is a constant, such as -1 from 5 to 6, the function has a constant rate of change, forming a linear segment on the graph.

Critical Points

Critical points are very special points on the graph of a function where the derivative is either zero or undefined. These points signal where the function's behavior might change, such as switching from increasing to decreasing or vice versa.
In our exercise, we have critical points at \(x = 2\) and \(x = 4\). At these points, the derivative \(f'(x)\) is zero, meaning the slope of the tangent is horizontal. Critical points often correspond to peaks or valleys on the graph.

• If the function changes from increasing to decreasing at a critical point, it's likely a local maximum. For example, at \(x=2\), the graph reaches a high point.
• If it changes from decreasing to increasing, it's likely a local minimum.

Understanding critical points is crucial because they help pinpoint where the graph should change direction, affecting the overall shape of the curve.

Concavity

Concavity refers to the direction in which a curve bends. To determine the concavity of a function, we look at the second derivative, \(f''(x)\).
If \(f''(x) > 0\), the graph is concave up, resembling a cup shape where the function exhibits an upward bend. On the other hand, if \(f''(x) < 0\), the graph is concave down, looking like an upside-down cup or arching downward.
In our exercise, the function is:

• Concave down on the intervals \((0,3)\) and \((4,5)\).
• Concave up on the interval \((3,4)\).

The switch in concavity at \(x=3\) is known as an inflection point, where the graph changes its curvature direction. Understanding concavity is essential as it helps convey how the function's speed of change itself changes, further shaping the graph's appearance.