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Chapter 4: Problem 13

Use the first derivative to find all critical points and use the second derivative to find all inflection points. Use a graph to identify each critical point as a local maximum, a local minimum, or neither. \(f(x)=2 x^{3}+3 x^{2}-36 x+5\)

Short Answer

Expert verified
Critical points: \(x = -3\) (max), \(x = 2\) (min). Inflection point: \(x = -\frac{1}{2}\).

Step by step solution

01

Find the First Derivative

To find the critical points, we first need to find the derivative of the function. Given the function \( f(x) = 2x^3 + 3x^2 - 36x + 5 \), its first derivative is computed by using the power rule. Therefore, the derivative \( f'(x) \) is:\[ f'(x) = \frac{d}{dx}(2x^3 + 3x^2 - 36x + 5) = 6x^2 + 6x - 36. \]
02

Find Critical Points

Critical points occur where the first derivative is equal to zero or undefined. We solve the equation \( 6x^2 + 6x - 36 = 0 \) to find these points:\[ 6x^2 + 6x - 36 = 0. \]Dividing the entire equation by 6 simplifies it to \( x^2 + x - 6 = 0 \). This quadratic equation can be factored as:\[ (x - 2)(x + 3) = 0. \]Setting each factor equal to zero gives the critical points: \( x = 2 \) and \( x = -3 \).
03

First Derivative Test

Using a graph or test intervals, we determine whether each critical point is a local maximum or minimum. For \( x = -3 \) and \( x = 2 \), examine the sign of \( f'(x) \) in intervals around these points. The sign of the derivative indicates whether the function is increasing or decreasing.- For \( x < -3 \), choose \( x = -4 \), and compute \( f'(-4) = 30 \) (positive, increasing).- For \( -3 < x < 2 \), choose \( x = 0 \), and compute \( f'(0) = -36 \) (negative, decreasing).- For \( x > 2 \), choose \( x = 3 \), and compute \( f'(3) = 36 \) (positive, increasing).Thus, \( x = -3 \) is a local maximum, and \( x = 2 \) is a local minimum.
04

Find the Second Derivative

To find inflection points, we use the second derivative. Taking the derivative of \( f'(x) \), we get:\[ f''(x) = \frac{d}{dx}(6x^2 + 6x - 36) = 12x + 6. \]
05

Find Inflection Points

Inflection points occur where the second derivative changes sign, which can happen where it is zero or undefined. Solve \( 12x + 6 = 0 \):\[ 12x + 6 = 0 \]\[ 12x = -6 \]\[ x = -\frac{1}{2}. \]Check for a sign change in \( f''(x) \) around \( x = -\frac{1}{2} \) to confirm it's an inflection point. For \( x < -\frac{1}{2} \), choose \( x = -1 \): \( f''(-1) = -6 \) (negative). For \( x > -\frac{1}{2} \), choose \( x = 0 \): \( f''(0) = 6 \) (positive). This confirms a sign change.
06

Graph the Function

Graph \( f(x) = 2x^3 + 3x^2 - 36x + 5 \) and note the critical and inflection points. The critical points at \( x = -3 \) (local maximum) and \( x = 2 \) (local minimum) should be evident on the graph. The curve will change concavity around the inflection point at \( x = -\frac{1}{2} \).

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Key Concepts

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

First Derivative
A key element in calculus is finding the first derivative of a function, which represents the function's rate of change or slope. For a given function, such as \[ f(x) = 2x^3 + 3x^2 - 36x + 5 \],we apply the power rule to find its derivative. This allows us to determine where the function is increasing or decreasing. In this exercise, the first derivative \[ f'(x) = 6x^2 + 6x - 36 \]is computed. Crucially, critical points occur where this derivative equals zero or is undefined. Solving \[ 6x^2 + 6x - 36 = 0 \]helps identify these points. It's a key step leading to understanding the nature of the critical points, which may be local maximums or minimums.
Second Derivative
The second derivative provides insights into the function's concavity and helps find inflection points. It is the derivative of the derivative. By analyzing \[ f''(x) = 12x + 6 \],you can identify where the concavity of the function changes. This happens when the second derivative equals zero. Solving \[ 12x + 6 = 0 \]gives us the x-coordinate \[ x = -\frac{1}{2} \].Verifying a change in sign of the second derivative around this x-value confirms the presence of an inflection point. The second derivative tells how the slope of a function's graph is changing, adding another layer to understanding its behavior.
Local Maximum and Minimum
Local maximums and minimums are the peaks and troughs on a function's graph. The first derivative test is used to classify critical points into local maximum or minimum. The sign of the first derivative changes when you examine points around the critical values.
  • For \( x = -3 \), the derivative changes from positive to negative, indicating a local maximum.
  • For \( x = 2 \), the change is from negative to positive, showing a local minimum.
Knowing whether the function is increasing or decreasing around these points helps determine their nature, providing a detailed view of the function's peaks and troughs.
Inflection Point
An inflection point is where the graph changes its concavity, from curved upwards to downwards or vice versa. To find it, solve where the function's second derivative equals zero. In our example,\[ x = -\frac{1}{2} \]is where the second derivative was zero, and it changed sign around this point. Before the inflection point, the derivative was negative, indicating a concave down curve. After it, the derivative turned positive, marking a concave up curve. Spotting an inflection point is crucial for capturing the graphical subtleties of a function.
Graphical Analysis of Functions
Visualizing functions through their graphs can make understanding complex calculus concepts easier. When graphing \[ f(x) = 2x^3 + 3x^2 - 36x + 5 \],you'll see critical points at \( x = -3 \) and \( x = 2 \), representing a local maximum and minimum, respectively. Additionally, at the inflection point \( x = -\frac{1}{2} \), the graph will change its concavity.

Observing these features on a graph provides a tangible view of abstract ideas, allowing you to appreciate the function鈥檚 behavior beyond numerical calculation. Graphical analysis thus equips you with a deeper understanding of a function's dynamics.

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Most popular questions from this chapter

School organizations raise money by selling candy door to door. The table shows \(p\), the price of the candy, and \(q\), the quantity sold at that price. $$\begin{array}{c|c|c|c|c|c|c|c} \hline p & \$ 1.00 & \$ 1.25 & \$ 1.50 & \$ 1.75 & \$ 2.00 & \$ 2.25 & \$ 2.50 \\\ \hline q & 2765 & 2440 & 1980 & 1660 & 1175 & 800 & 430 \\ \hline \end{array}$$ (a) Estimate the elasticity of demand at a price of $$\$ 1.00$$. At this price, is the demand elastic or inelastic? (b) Estimate the elasticity at each of the prices shown. What do you notice? Give an explanation for why this might be so. (c) At approximately what price is elasticity equal to 1 ? (d) Find the total revenue at each of the prices shown. Confirm that the total revenue appears to be maximized at approximately the price where \(E=1\).

The income elasticity of demand for a product is defined as \(E_{\text {income }}=|I / q \cdot d q / d I|\) where \(q\) is the quantity demanded as a function of the income \(I\) of the consumer. What does \(E_{\text {income }}\) tell you about the sensitivity of the quantity of the product purchased to changes in the income of the consumer?

A warehouse selling cement has to decide how often and in what quantities to reorder. It is cheaper, on average, to place large orders, because this reduces the ordering cost per unit. On the other hand, larger orders mean higher storage costs. The warehouse always reorders cement in the same quantity, \(q\). The total weekly cost, \(C\), of ordering and storage is given by \(C=\frac{a}{q}+b q, \quad\) where \(a, b\) are positive constants. (a) Which of the terms, \(a / q\) and \(b q\), represents the ordering cost and which represents the storage cost? (b) What value of \(q\) gives the minimum total cost?

As an epidemic spreads through a population, the number of infected people, \(I\), is expressed as a function of the number of susceptible people, \(S\), by \(I=k \ln \left(\frac{S}{S_{0}}\right)-S+S_{0}+I_{0}, \quad\) for \(k, S_{0}, I_{0}>0\) (a) Find the maximum number of infected people. (b) The constant \(k\) is a characteristic of the particular disease; the constants \(S_{0}\) and \(I_{0}\) are the values of \(S\) and \(I\) when the disease starts. Which of the following affects the maximum possible value of \(I ? \mathrm{Ex}-\) plain. \- The particular disease, but not how it starts. \- How the disease starts, but not the particular disease. \- Both the particular disease and how it starts.

The demand for a product is given by \(p=90-10 q\). Find the elasticity of demand when \(p=50\). If this price rises by \(2 \%\), calculate the corresponding percentage change in demand.
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