The Concept of Decreasing Functions

Oct 13, 2025 by windailys.com 46 views

Introduction

In calculus, the study of functions and their behavior is crucial in understanding various mathematical concepts. One of the key aspects of function analysis is determining the intervals on which a function is increasing or decreasing. This is particularly important in optimization problems, where we need to find the maximum or minimum value of a function. In this article, we will focus on finding the intervals on which the function $g(x) = 1 - x^3$ is decreasing.

The Concept of Decreasing Functions

A function $f(x)$ is said to be decreasing on an interval $(a, b)$ if for any two points $x_1$ and $x_2$ in the interval, $f(x_1) > f(x_2)$ whenever $x_1 < x_2$ . In other words, the function decreases as the input value increases. To determine the intervals on which a function is decreasing, we need to find the critical points of the function, which are the points where the function changes from increasing to decreasing or vice versa.

Finding the Critical Points

To find the critical points of the function $g(x) = 1 - x^3$ , we need to find the points where the derivative of the function is equal to zero or undefined. The derivative of the function is given by:

g'(x) = -3x^2

Setting the derivative equal to zero, we get:

-3x^2 = 0

Solving for $x$ , we get:

x = 0

This is the only critical point of the function.

Determining the Intervals of Decrease

To determine the intervals on which the function is decreasing, we need to examine the behavior of the function on either side of the critical point. We can do this by evaluating the derivative of the function at a point to the left and right of the critical point.

Let's evaluate the derivative at $x = -1$ :

g'(-1) = -3(-1)^2 = -3

Since the derivative is negative, the function is decreasing at $x = -1$ .

Now, let's evaluate the derivative at $x = 1$ :

g'(1) = -3(1)^2 = -3

Since the derivative is negative, the function is decreasing at $x = 1$ .

Conclusion

Based on the analysis above, we can conclude that the function $g(x) = 1 - x^3$ is decreasing on the intervals $(-\infty, 0)$ and $(0, \infty)$ .

The Sign of the Derivative

To confirm our conclusion, let's examine the sign of the derivative on the intervals $(-\infty, 0)$ and $(0, \infty)$ .

On the interval $(-\infty, 0)$ , the derivative is negative, since $-3x^2 < 0$ for all $x < 0$ . Therefore, the function is decreasing on this interval.

On the interval $(0, \infty)$ , the derivative is negative, since $-3x^2 < 0$ for all $x > 0$ . Therefore, the function is decreasing on this interval.

The Intervals of Decrease

Based on the analysis above, we can conclude that the function $g(x) = 1 - x^3$ is decreasing on the intervals $(-\infty, 0)$ and $(0, \infty)$ .

The Graph of the Function

The graph of the function $g(x) = 1 - x^3$ is a cubic curve that opens downward. The function has a critical point at $x = 0$ , where the function changes from increasing to decreasing.

The Behavior of the Function

The function $g(x) = 1 - x^3$ is a decreasing function on the intervals $(-\infty, 0)$ and $(0, \infty)$ . The function has a critical point at $x = 0$ , where the function changes from increasing to decreasing.

The Sign of the Second Derivative

To confirm our conclusion, let's examine the sign of the second derivative of the function.

The second derivative of the function is given by:

g''(x) = -6x

Evaluating the second derivative at $x = 0$ , we get:

g''(0) = 0

Since the second derivative is zero at the critical point, we need to examine the sign of the second derivative on either side of the critical point.

On the interval $(-\infty, 0)$ , the second derivative is negative, since $-6x < 0$ for all $x < 0$ . Therefore, the function is concave downward on this interval.

On the interval $(0, \infty)$ , the second derivative is negative, since $-6x < 0$ for all $x > 0$ . Therefore, the function is concave downward on this interval.

The Conclusion

Based on the analysis above, we can conclude that the function $g(x) = 1 - x^3$ is decreasing on the intervals $(-\infty, 0)$ and $(0, \infty)$ .

The Final Answer

Q: What is the function $g(x) = 1 - x^3$ ?

A: The function $g(x) = 1 - x^3$ is a cubic function that opens downward. It has a critical point at $x = 0$ , where the function changes from increasing to decreasing.

Q: What is the derivative of the function $g(x) = 1 - x^3$ ?

A: The derivative of the function $g(x) = 1 - x^3$ is given by:

g'(x) = -3x^2

Q: How do you find the critical points of the function $g(x) = 1 - x^3$ ?

A: To find the critical points of the function $g(x) = 1 - x^3$ , we need to find the points where the derivative of the function is equal to zero or undefined. In this case, the derivative is equal to zero at $x = 0$ .

Q: What is the behavior of the function $g(x) = 1 - x^3$ on the intervals $(-\infty, 0)$ and $(0, \infty)$ ?

A: The function $g(x) = 1 - x^3$ is decreasing on the intervals $(-\infty, 0)$ and $(0, \infty)$ . The function has a critical point at $x = 0$ , where the function changes from increasing to decreasing.

Q: How do you determine the intervals on which a function is decreasing?

A: To determine the intervals on which a function is decreasing, we need to examine the behavior of the function on either side of the critical point. We can do this by evaluating the derivative of the function at a point to the left and right of the critical point.

Q: What is the significance of the second derivative of the function $g(x) = 1 - x^3$ ?

A: The second derivative of the function $g(x) = 1 - x^3$ is given by:

g''(x) = -6x

The second derivative is negative on the intervals $(-\infty, 0)$ and $(0, \infty)$ , indicating that the function is concave downward on these intervals.

Q: What is the final answer to the problem?

A: The final answer is $\boxed{(-\infty, 0) \cup (0, \infty)}$ .

Q: What is the graph of the function $g(x) = 1 - x^3$ ?

A: The graph of the function $g(x) = 1 - x^3$ is a cubic curve that opens downward. The function has a critical point at $x = 0$ , where the function changes from increasing to decreasing.

Q: What is the behavior of the function $g(x) = 1 - x^3$ at the critical point $x = 0$ ?

A: The function $g(x) = 1 - x^3$ is decreasing at the critical point $x = 0$ . The function changes from increasing to decreasing at this point.

Q: How do you use the derivative and second derivative to analyze the behavior of the function $g(x) = 1 - x^3$ ?

A: We can use the derivative and second derivative to analyze the behavior of the function $g(x) = 1 - x^3$ by examining the sign of the derivative and second derivative on different intervals. This will help us determine the intervals on which the function is increasing or decreasing.

Q: What is the importance of analyzing the behavior of a function?

A: Analyzing the behavior of a function is important in understanding various mathematical concepts, such as optimization problems. It also helps us to determine the intervals on which a function is increasing or decreasing, which is crucial in many real-world applications.