Finding Critical Points Of F(x) = X(4-x)^3 A Step-by-Step Guide

Jul 15, 2025 by ADMIN 64 views

Introduction

In calculus, critical points play a vital role in analyzing the behavior of functions. These points, where the derivative is either zero or undefined, are crucial for identifying local maxima, local minima, and saddle points. Understanding how to determine critical points is essential for sketching graphs, solving optimization problems, and gaining a deeper insight into the nature of a function. In this article, we will walk through the process of finding critical points for the function $f(x) = x(4-x)^3$ . This involves applying the product rule and chain rule to find the derivative, setting the derivative equal to zero, and solving for $x$ . By the end of this guide, you will have a clear understanding of how to find and interpret critical points for similar functions.

Understanding Critical Points

To begin, let's define what critical points are and why they matter. Critical points of a function $f(x)$ are the $x$ -values where the derivative $f'(x)$ is either equal to zero or undefined. These points are significant because they indicate where the function's slope changes direction, potentially leading to local maxima, local minima, or points of inflection. A local maximum is a point where the function's value is higher than all nearby points, while a local minimum is a point where the function's value is lower than all nearby points. Saddle points, on the other hand, are points where the function's slope changes, but it is neither a maximum nor a minimum. Critical points are the foundation for many calculus applications, including optimization problems, curve sketching, and the analysis of function behavior. For example, in optimization, critical points help us find the maximum or minimum values of a function, which can be applied in various fields such as engineering, economics, and physics. In curve sketching, knowing the critical points allows us to identify key features of a function's graph, such as peaks, valleys, and flat regions. Understanding and finding critical points is a fundamental skill in calculus, providing a pathway to deeper insights into functions and their applications.

Step-by-Step Solution for f(x) = x(4-x)^3

1. Find the Derivative f'(x)

The first step in finding the critical points of $f(x) = x(4-x)^3$ is to compute its derivative, $f'(x)$ . To do this, we will apply the product rule and the chain rule. The product rule states that if we have a function $f(x) = u(x)v(x)$ , then $f'(x) = u'(x)v(x) + u(x)v'(x)$ . In our case, let $u(x) = x$ and $v(x) = (4-x)^3$ . The derivative of $u(x)$ is simply $u'(x) = 1$ . For $v(x)$ , we need to use the chain rule. The chain rule states that if we have a composite function $v(x) = g(h(x))$ , then $v'(x) = g'(h(x))h'(x)$ . Here, we can think of $g(y) = y^3$ and $h(x) = 4-x$ . Thus, $g'(y) = 3y^2$ and $h'(x) = -1$ . Applying the chain rule, we get $v'(x) = 3(4-x)^2(-1) = -3(4-x)^2$ . Now, we can apply the product rule:

$f'(x) = u'(x)v(x) + u(x)v'(x) = 1 imes (4-x)^3 + x imes (-3(4-x)^2)$ .

This simplifies to:

$f'(x) = (4-x)^3 - 3x(4-x)^2$ .

2. Simplify the Derivative

After finding the derivative, the next step is to simplify the expression. This makes it easier to find the roots, which are the $x$ -values where $f'(x) = 0$ . To simplify $f'(x) = (4-x)^3 - 3x(4-x)^2$ , we can factor out the common term $(4-x)^2$ :

$f'(x) = (4-x)^2[(4-x) - 3x]$ .

Now, simplify the expression inside the brackets:

$f'(x) = (4-x)^2(4 - x - 3x) = (4-x)^2(4 - 4x)$ .

Further simplification involves factoring out 4 from the second term:

$f'(x) = 4(4-x)^2(1 - x)$ .

This simplified form of the derivative is much easier to work with when finding critical points.

3. Set f'(x) = 0 and Solve for x

To find the critical points, we need to determine the values of $x$ for which $f'(x) = 0$ . Using the simplified derivative $f'(x) = 4(4-x)^2(1 - x)$ , we set it equal to zero:

$4(4-x)^2(1 - x) = 0$ .

This equation is satisfied if any of the factors are equal to zero. Thus, we have two cases to consider:

$(4-x)^2 = 0$
$(1-x) = 0$

For the first case, $(4-x)^2 = 0$ , taking the square root of both sides gives:

$4 - x = 0$ ,

which implies:

$x = 4$ .

For the second case, $(1-x) = 0$ , solving for $x$ gives:

$x = 1$ .

Therefore, the solutions to $f'(x) = 0$ are $x = 1$ and $x = 4$ . These are the $x$ -coordinates of the critical points of the function $f(x) = x(4-x)^3$ .

4. Identify Critical Points

Now that we have found the $x$ -values where $f'(x) = 0$ , we need to identify the critical points. The critical points occur at $x=1$ and $x=4$ . These points are where the function's slope might change direction, indicating potential local maxima, minima, or saddle points. In this case, we found two critical points: $x=1$ and $x=4$ . To fully understand the nature of these critical points, we would typically perform additional analysis, such as using the first derivative test or the second derivative test. The first derivative test involves examining the sign of $f'(x)$ in the intervals around the critical points to determine whether the function is increasing or decreasing. The second derivative test involves evaluating the second derivative $f''(x)$ at the critical points to determine the concavity of the function. In summary, we have successfully identified the critical points for the function $f(x) = x(4-x)^3$ by finding the values of $x$ where the derivative $f'(x)$ is equal to zero.

Importance of Critical Points

Critical points are fundamental in calculus for several reasons. They help in identifying the local maxima and minima of a function, which are crucial for optimization problems. For instance, in engineering, finding the minimum cost or maximum efficiency often involves identifying critical points of a cost or performance function. In economics, critical points can help determine the production level that maximizes profit. Understanding the behavior of a function near its critical points is also essential for sketching its graph. By analyzing the sign of the first derivative around the critical points, we can determine where the function is increasing or decreasing. Similarly, the second derivative test can help us identify the concavity of the function, indicating whether it is curving upwards or downwards. This information is invaluable for creating an accurate and informative graph of the function. Moreover, critical points are used in various mathematical and scientific applications, including physics (finding equilibrium points), statistics (finding maximum likelihood estimators), and computer science (optimization algorithms). The concept of critical points is a cornerstone of calculus, providing essential tools for analyzing and understanding functions in a wide range of contexts. Knowing how to find and interpret critical points is a crucial skill for anyone working with mathematical models and quantitative analysis.

Conclusion

In this article, we have thoroughly explored the process of finding critical points for the function $f(x) = x(4-x)^3$ . We began by understanding the definition and significance of critical points, which are the $x$ -values where the derivative of a function is either zero or undefined. These points are crucial for identifying local maxima, local minima, and saddle points, and they play a vital role in various applications, including optimization, curve sketching, and analysis of function behavior. We then walked through a step-by-step solution to find the critical points for our given function. This involved applying the product rule and chain rule to compute the derivative, simplifying the derivative expression, setting the derivative equal to zero, and solving for $x$ . We found that the critical points for $f(x) = x(4-x)^3$ occur at $x = 1$ and $x = 4$ . We also discussed the broader importance of critical points in calculus and their applications in various fields. Critical points are not just mathematical concepts; they are powerful tools that provide insights into the behavior and properties of functions. By mastering the techniques for finding and interpreting critical points, you can enhance your understanding of calculus and its applications in real-world problems. The ability to analyze functions and identify critical points is a valuable skill for anyone working in mathematics, science, engineering, economics, and other quantitative disciplines.