The Instantaneous Rate of Change

Oct 13, 2025 by talynt.org 86 views

Introduction

In calculus, the instantaneous rate of change and the average rate of change are two fundamental concepts used to describe the behavior of functions. The instantaneous rate of change is a measure of the rate at which a function changes at a specific point, while the average rate of change is a measure of the rate at which a function changes over a given interval. In this article, we will explore the relationship between these two concepts and determine the value of $x$ in the closed interval $[0,5]$ for which the instantaneous rate of change of the function $f(x) = x^3 - 2x^2 + 5x - 16$ equals the average rate of change.

The Instantaneous Rate of Change

The instantaneous rate of change of a function $f(x)$ at a point $x=a$ is given by the derivative of the function at that point, denoted as $f'(a)$ . In other words, it is the rate at which the function changes at a specific point. To find the instantaneous rate of change of the function $f(x) = x^3 - 2x^2 + 5x - 16$ , we need to find its derivative.

Finding the Derivative

To find the derivative of the function $f(x) = x^3 - 2x^2 + 5x - 16$ , we will use the power rule of differentiation, which states that if $f(x) = x^n$ , then $f'(x) = nx^{n-1}$ . Applying this rule to each term of the function, we get:

f'(x) = \frac{d}{dx}(x^3 - 2x^2 + 5x - 16)

f'(x) = 3x^2 - 4x + 5

The Average Rate of Change

The average rate of change of a function $f(x)$ over an interval $[a,b]$ is given by the formula:

\frac{f(b) - f(a)}{b - a}

In this case, we are interested in finding the average rate of change of the function $f(x) = x^3 - 2x^2 + 5x - 16$ over the interval $[0,5]$ .

Finding the Average Rate of Change

To find the average rate of change of the function $f(x) = x^3 - 2x^2 + 5x - 16$ over the interval $[0,5]$ , we need to evaluate the function at the endpoints of the interval and then apply the formula for the average rate of change.

f(0) = (0)^3 - 2(0)^2 + 5(0) - 16 = -16

f(5) = (5)^3 - 2(5)^2 + 5(5) - 16 = 125 - 50 + 25 - 16 = 84

Now, we can apply the formula for the average rate of change:

\frac{f(5) - f(0)}{5 - 0} = \frac{84 - (-16)}{5} = \frac{100}{5} = 20

Equating the Instantaneous and Average Rates of Change

Now that we have found the instantaneous rate of change of the function $f(x) = x^3 - 2x^2 + 5x - 16$ and the average rate of change over the interval $[0,5]$ , we can equate the two expressions and solve for $x$ .

f'(x) = 3x^2 - 4x + 5 = 20

3x^2 - 4x - 15 = 0

To solve this quadratic equation, we can use the quadratic formula:

x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

In this case, $a = 3$ , $b = -4$ , and $c = -15$ . Plugging these values into the formula, we get:

x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(3)(-15)}}{2(3)}

x = \frac{4 \pm \sqrt{16 + 180}}{6}

x = \frac{4 \pm \sqrt{196}}{6}

x = \frac{4 \pm 14}{6}

Simplifying the expression, we get two possible values for $x$ :

x = \frac{4 + 14}{6} = \frac{18}{6} = 3

x = \frac{4 - 14}{6} = \frac{-10}{6} = -\frac{5}{3}

However, we are interested in the value of $x$ that lies within the closed interval $[0,5]$ . Therefore, we can discard the value $x = -\frac{5}{3}$ and conclude that the value of $x$ for which the instantaneous rate of change of the function $f(x) = x^3 - 2x^2 + 5x - 16$ equals the average rate of change over the interval $[0,5]$ is $x = 3$ .

Conclusion

Introduction

In our previous article, we explored the relationship between the instantaneous rate of change and the average rate of change of a function. We found the instantaneous rate of change of the function $f(x) = x^3 - 2x^2 + 5x - 16$ and the average rate of change over the interval $[0,5]$ . We then equated the two expressions and solved for $x$ to find the value of $x$ for which the instantaneous rate of change equals the average rate of change. In this article, we will answer some frequently asked questions related to the instantaneous and average rates of change.

Q&A

Q: What is the difference between the instantaneous rate of change and the average rate of change?

A: The instantaneous rate of change is a measure of the rate at which a function changes at a specific point, while the average rate of change is a measure of the rate at which a function changes over a given interval.

Q: How do you find the instantaneous rate of change of a function?

A: To find the instantaneous rate of change of a function, you need to find its derivative. The derivative of a function $f(x)$ is denoted as $f'(x)$ and represents the rate at which the function changes at a specific point.

Q: How do you find the average rate of change of a function?

A: To find the average rate of change of a function, you need to evaluate the function at the endpoints of the interval and then apply the formula:

\frac{f(b) - f(a)}{b - a}

Q: Can the instantaneous rate of change and the average rate of change be equal?

A: Yes, the instantaneous rate of change and the average rate of change can be equal. This occurs when the function is linear, meaning that its derivative is constant over the interval.

Q: How do you solve for $x$ when the instantaneous rate of change equals the average rate of change?

A: To solve for $x$ when the instantaneous rate of change equals the average rate of change, you need to equate the two expressions and solve for $x$ . This may involve using algebraic techniques, such as factoring or the quadratic formula.

Q: What is the significance of the instantaneous rate of change and the average rate of change in real-world applications?

A: The instantaneous rate of change and the average rate of change have many real-world applications, such as modeling population growth, predicting stock prices, and analyzing the behavior of physical systems.

Q: Can the instantaneous rate of change and the average rate of change be used to determine the maximum or minimum value of a function?

A: Yes, the instantaneous rate of change and the average rate of change can be used to determine the maximum or minimum value of a function. This is because the derivative of a function represents the rate at which the function changes, and the sign of the derivative can indicate whether the function is increasing or decreasing.

Conclusion

In this article, we have answered some frequently asked questions related to the instantaneous and average rates of change. We have discussed the difference between the instantaneous rate of change and the average rate of change, how to find the instantaneous rate of change and the average rate of change, and how to solve for $x$ when the instantaneous rate of change equals the average rate of change. We have also discussed the significance of the instantaneous rate of change and the average rate of change in real-world applications and how they can be used to determine the maximum or minimum value of a function.

Additional Resources

For more information on the instantaneous and average rates of change, we recommend the following resources:

Calculus textbooks, such as "Calculus" by Michael Spivak or "Calculus: Early Transcendentals" by James Stewart
Online resources, such as Khan Academy or MIT OpenCourseWare
Research articles and papers on the topic of instantaneous and average rates of change

Glossary

Instantaneous rate of change: A measure of the rate at which a function changes at a specific point.
Average rate of change: A measure of the rate at which a function changes over a given interval.
Derivative: The rate at which a function changes at a specific point, denoted as $f'(x)$ .
Quadratic formula: A formula used to solve quadratic equations, given by:

x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Index

Calculus
Derivative
Instantaneous rate of change
Average rate of change
Quadratic formula
Real-world applications