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Chapter 4: Problem 218

Draw a graph that satisfies the given specifications for the domain \(x=[-3,3]\). The function does not have to be continuous or differentiable. \(\quad f^{\prime \prime}(x)<0 \quad\) over \(-10,-3

Short Answer

Expert verified
Draw a curve with concavity constraints, maximizing at \(x=0\) and minimizing at \(x = \pm 2\).

Step by step solution

01

Understand the Given Specifications

First, we need to understand what the given specifications mean. The graph must have: \( f''(x) < 0\) for \(-1 < x < 1\) which means the graph is concave down in this interval. \( f''(x) > 0\) when \(-3 < x < -1\) and \(1 < x < 3\), so the graph is concave up in these intervals. There should be a local maximum at \(x=0\) and local minima at \(x=\pm 2\).
02

Sketch Basic Structure Based on Concavity

Begin by sketching the basic structure of the function using the intervals of concavity. \(f''(x) < 0\) for \(-1 < x < 1\) suggests a shape like an upside-down bowl, and \(f''(x) > 0\) for \(-3 < x < -1\) and \(1 < x < 3\) means right-side-up bowls. This helps place the regions where the curve bends.
03

Place Local Maximum and Minima

Place a local maximum at \(x = 0\). This will be the peak of the upside-down bowl. At \(x = -2\) and \(x = 2\), place local minima. Both these points will be the bottoms of the right-side-up bowls. Ensure the changes between these points reflect the concavity given.
04

Connect Points Considering Domain and Cues

Now, start connecting these points smoothly within the specified domain \([-3, 3]\) based on the concavity. Ensure the curve transitions smoothly between \(x = -3\) and \(x = -1\), \(x = -1\) to \(x = 1\), and \(x = 1\) to \(x = 3\) while respecting the concavity and critical points (local max and mins).
05

Verify the Graph Satisfies All Conditions

Lastly, check if the sketched graph satisfies all conditions: - It should be concave down in \(-1 < x < 1\), indicating \(f''(x) < 0\).- It should be concave up in \(-3 < x < -1\) and \(1 < x < 3\), indicating \(f''(x) > 0\).- Local maxima and minima should be correctly placed at \(x = 0\) and \(x = \pm 2\) respectively.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Concavity
Concavity reveals the direction in which a curve bends. Imagine the curve like a bowl. If the bowl opens upwards, it is 'concave up', and if it opens downwards, it's 'concave down'.
  • A graph is concave up if its derivative, which is the rate of change, is increasing. This happens when the second derivative, denoted as \( f''(x) \), is greater than zero.
  • Conversely, a graph is concave down if its derivative is decreasing. This occurs when \( f''(x) < 0 \), indicating the curve is bending downwards.
      • When sketching a curve, the intervals of concavity help determine the overall shape. For example, in the domain \([-3,3]\) with \( f''(x) < 0 \) for \(-1 < x < 1\), the curve is concave down, resembling an upside-down bowl. Alternatively, for \( f''(x) > 0 \) in the intervals \(-3 < x < -1\) and \(+1 < x < 3\), the curve is concave up, appearing like a right-side-up bowl. This behavior influences where peaks and valleys occur on the graph.
Identifying Local Maximum
A local maximum is a point where the function reaches a peak. This means that the value of the function at this point is higher than at any other point in the immediate vicinity.
  • To identify a local maximum, we look for where the first derivative, \( f'(x) \), changes from positive to negative. This implies a rise followed by a fall, akin to a peak on a hill.
  • Mathematically, it occurs where \( f'(x) = 0 \) and \( f''(x) < 0 \).
For a curve subject to our exercise's conditions, the local maximum is specified at \( x = 0 \). This point represents the top of the upside-down bowl within our specified range. Evaluating the change in sign of the first derivative helps confirm the positioning of this local maximum.
Spotting Local Minimum
Local minima are the troughs in a graph, where the function takes its lowest value in a particular region.
  • A local minimum is characterized by the derivative changing from negative to positive, indicating a drop followed by an ascent.
  • It occurs when \( f'(x) = 0 \) with \( f''(x) > 0 \).
In the problem, local minima are located at \( x = -2 \) and \( x = 2 \). These points match the bottoms of the right-side-up bowls. Understanding the transitions at these intervals ensures accurate plotting of these local minima on the graph.
Roles of Derivatives
Derivatives play a crucial role in understanding the behavior of functions. They help determine the rate at which one quantity changes in relation to another.
  • The first derivative, \( f'(x) \), provides information about the slope of the function. When \( f'(x) > 0 \), the function increases; when \( f'(x) < 0 \), it decreases.
  • The second derivative, \( f''(x) \), reveals concavity, offering insights into how the curve bends.
In our example, we use derivatives to pinpoint points of interest, such as maxima and minima, and to understand how to draw the graph according to the curvature rules set by the second derivative. By correctly interpreting these derivatives, the construction of an accurate curve adhering to the problem's requirements is made possible.

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Most popular questions from this chapter

For the following exercises, determine whether the statement is true or false. Either prove it is true or find a counterexample if it is false. If \(f(x)\) is the antiderivative of \(v(x),\) then \(f(x)+1\) is the antiderivative of \(v(x)+1\) .

For the following exercises, determine whether the statement is true or false. Either prove it is true or find a counterexample if it is false. If \(f(x)\) is the antiderivative of \(v(x),\) then \((f(x))^{2}\) is the antiderivative of \((v(x))^{2}\) .

For the following exercises, consider the points on the given graphs. Use a calculator to graph the functions. Where is the line \(y=5-2 x\) closest to the origin?

For the following exercises, use both Newton鈥檚 method and the secant method to calculate a root for the following equations. Use a calculator or computer to calculate how many iterations of each are needed to reach within three decimal places of the exact answer. For the secant method, use the first guess from Newton鈥檚 method. $$f(x)=e^{x}-1, x_{0}=2$$

The following two exercises consider a bank investment. The initial investment is \(\$ 10,000\) . After 25 years, the investment has tripled to \(\$ 30,000\) . Use Newton鈥檚 method to determine the interest rate if the interest was compounded continuously.
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