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Estimating rates of change: Use your calculator to make the graph of \(f(x)=x^{3}-5 x\). a. Is \(\frac{d f}{d x}\) positive or negative at \(x=2\) ? b. Identify a point on the graph of \(f\) where \(\frac{d f}{d x}\) is negative.

Short Answer

Expert verified
a. The derivative is positive at \( x=2 \). b. At \( x=0 \), the derivative is negative.

Step by step solution

01

Understanding the Derivative

To determine if the function \( f(x) = x^3 - 5x \) is increasing or decreasing at a given point, we need to find its derivative, \( \frac{df}{dx} \). The derivative gives the rate of change of the function at any point \( x \).
02

Calculate the Derivative

Use the power rule for derivatives to find \( \frac{df}{dx} \). The derivative of \( f(x) = x^3 - 5x \) is \( \frac{df}{dx} = 3x^2 - 5 \). This tells us how the function changes at each point \( x \).
03

Evaluate the Derivative at x = 2

Plug \( x = 2 \) into the derivative to check if the rate of change is positive or negative. \( \frac{df}{dx} \vert_{x=2} = 3(2)^2 - 5 = 12 - 5 = 7 \), which is positive. Therefore, the function is increasing at \( x = 2 \).
04

Identifying a Point where the Derivative is Negative

Find a value of \( x \) such that \( \frac{df}{dx} = 3x^2 - 5 < 0 \). Set \( 3x^2 < 5 \), then \( x^2 < \frac{5}{3} \), so \( -\sqrt{\frac{5}{3}} < x < \sqrt{\frac{5}{3}} \). Approximately, \( x \) must be between \( -1.29 \) and \( 1.29 \) to have a negative derivative.
05

Verify a Specific Point

Choose a specific value within the range \( -1.29 < x < 1.29 \). Check \( x = 0 \), then \( \frac{df}{dx} \vert_{x=0} = 3(0)^2 - 5 = -5 \). This is negative, confirming the function decreases at \( x = 0 \).

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Key Concepts

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

Derivative
The derivative is a fundamental concept in calculus, representing the rate at which a function changes at any given point. Calculating the derivative of a function involves finding how the output value of the function changes with a slight change in the input value. It provides crucial information about the behavior of functions, such as whether they are increasing or decreasing.

To find the derivative, we often use rules like the power rule, the product rule, or the chain rule, depending on the complexity of the function. In the case of the function \( f(x) = x^3 - 5x \), its derivative is found using the power rule, leading to \( \frac{df}{dx} = 3x^2 - 5 \). This derivative \( \frac{df}{dx} \) tells us the rate of change at any point \( x \) along the curve.
Rate of Change
The rate of change is a measure of how quickly a function's value changes as the input changes. It is closely tied to the concept of the derivative in calculus.

When we calculate the derivative of a function, we actually determine the instantaneous rate of change at a specific point. For example, evaluating the derivative \( \frac{df}{dx} = 3x^2 - 5 \) at \( x = 2 \) gives us a rate of change of \( 7 \). This positive value indicates that at \( x = 2 \), the function \( f(x) = x^3 - 5x \) is increasing. Understanding the rate of change allows us to predict the behavior of functions and graphs, determine trends, and solve practical problems.
Increasing and Decreasing Functions
A function is increasing on an interval if, as \( x \) increases, the function value \( f(x) \) also increases. Conversely, a function is decreasing if \( f(x) \) decreases as \( x \) increases within an interval.

To determine whether a function is increasing or decreasing, we examine its derivative. If \( \frac{df}{dx} > 0 \) for a certain interval, the function is increasing within that region. If \( \frac{df}{dx} < 0 \), the function is decreasing. For example, for \( f(x) = x^3 - 5x \), the derivative \( 3x^2 - 5 \) is positive at \( x = 2 \), meaning the function is increasing at this point.
  • When \( x \) ranges between approximately \( -1.29 \) and \( 1.29 \), the derivative \( 3x^2 - 5 \) is negative, indicating that the function decreases in this interval.
By analyzing these changes, we discover where the function rises or falls and make informed predictions or interpretations accordingly.
Critical Points
Critical points occur where the derivative of a function is zero or undefined. These points are vital in determining the function's behavior, as they often signal a change from increasing to decreasing (or vice versa) or indicate points where the function reaches local maxima or minima.

To find critical points of \( f(x) = x^3 - 5x \), set the derivative \( \frac{df}{dx} = 3x^2 - 5 \) to zero and solve for \( x \). Solving \( 3x^2 - 5 = 0 \) gives us the critical points at \( x = \pm\sqrt{\frac{5}{3}} \). By examining these critical points, we can further analyze the behavior of the function. In these regions, the function may transition from increasing to decreasing or vice versa, providing crucial insights into the curve's overall shape.
Understanding critical points helps in sketching the graph of a function and predicting points of interest, such as peaks and troughs in the graph.

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Most popular questions from this chapter

11\. A population of bighorn sheep: There is an effort in Colorado to restore the population of bighorn sheep. Let \(N=N(t)\) denote the number of sheep in a certain protected area at time \(t\). a. Explain the meaning of \(\frac{d N}{d t}\) in practical terms. b. A small breeding population of bighorn sheep is initially introduced into the protected area. Food is plentiful and conditions are generally favorable for bighorn sheep. What would you expect to be true about the sign of \(\frac{d N}{d t}\) during this period? c. This summer a number of dead sheep were discovered, and all were infected with a disease that is known to spread rapidly among bighorn sheep and is nearly always fatal. How would you expect an unchecked spread of this disease to affect \(\frac{d N}{d t}\) ? d. If the reintroduction program goes well, then the population of bighorn sheep will grow to the size the available food supply can support and will remain at about that same level. What would you expect to be true of \(\frac{d N}{d t}\) when this happens?

Hiking: You are hiking in a hilly region, and \(E=\) \(E(t)\) is your elevation at time \(t\). a. Explain the meaning of \(\frac{d E}{d t}\) in practical terms. b. Where might you be when \(\frac{d E}{d t}\) is a large positive number? c. You reach a point where \(\frac{d E}{d t}\) is briefly zero. Where might you be? d. Where might you be when \(\frac{d E}{d t}\) is a large negative number?

Logistic growth with a threshold: Most species have a survival threshold level, and populations of fewer individuals than the threshold cannot sustain themselves. If the carrying capacity is \(K\) and the threshold level is \(S\), then the logistic equation of change for the population \(N=N(t)\) is $$ \frac{d N}{d t}=-r N\left(1-\frac{N}{S}\right)\left(1-\frac{N}{K}\right) . $$ For Pacific sardines, we may use \(K=2.4\) million tons and \(r=0.338\) per year, as in Example 6.10. Suppose we also know that the survival threshold level for the sardines is \(S=0.8\) million tons. a. Write the equation of change for Pacific sardines under these conditions. b. Make a graph of \(\frac{d N}{d t}\) versus \(N\) and use it to find the equilibrium solutions. How do the equilibrium solutions correspond with \(S\) and \(K\) ? c. For what values of \(N\) is the graph of \(N\) versus \(t\) increasing, and for what values is it decreasing? d. Explain what can be expected to happen to a population of \(0.7\) million tons of sardines. e. At what population level will the population be growing at its fastest?

Fishing for sardines: This is a continuation of Example 6.10. If we take into account an annual fish harvest of \(F\) million tons of fish, then the equation of change for Pacific sardines becomes $$ \frac{d N}{d t}=0.338 N\left(1-\frac{N}{2.4}\right)-F . $$ a. Suppose that there are currently \(1.8\) million tons of Pacific sardines off the California coast and that you are in charge of the commercial fishing fleet. It is your goal to leave the Pacific population of sardines as you found it. That is, you wish to set the fishing level \(F\) so that the biomass of Pacific sardines remains stable. What value of \(F\) will accomplish this? (Hint: You want to choose \(F\) so that the current biomass level of \(1.8\) million tons is an equilibrium solution.) b. For the remainder of this exercise, take the value of \(F\) to be \(0.1\) million tons per year. That is, assume the catch is 100,000 tons per year. i. Make a graph of \(\frac{d N}{d t}\) versus \(N\), and use it to find the equilibrium solutions. ii. For what values of \(N\) will the biomass be increasing? For what values will it be decreasing? iii. On the same graph, sketch all equilibrium solutions and the graphs of \(N\) versus \(t\) for each of the initial populations \(N(0)=0.3\) million tons, \(N(0)=1.0\) million tons, and \(N(0)=2.3\) million tons. \(\rightarrow\) iv. Explain in practical terms what the picture you made in part iii tells you. Include in your explanation the significance of the equilibrium solutions.

Growth of fish: Let \(w=w(t)\) denote the weight of a fish as a function of its age \(t\). For the North Sea cod, the equation of change $$ \frac{d w}{d t}=2.1 w^{2 / 3}-0.6 w $$ holds. Here \(w\) is measured in pounds and \(t\) in years. b. Make a graph of \(\frac{d w}{d t}\) against \(w\). Include weights up to 45 pounds. c. What is the weight of the cod when it is growing at the greatest rate? d. To what weight does the cod grow?

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