Nature of the Critical Points

Oct 13, 2025 by sorusgentrification.com 61 views

Introduction

In mathematics, a critical point of a function is a point where the derivative of the function is zero or undefined. These points are significant because they can indicate the presence of a maximum, minimum, or saddle point on the graph of the function. In this article, we will analyze the function $g(x) = x \sqrt{3-x}$ and determine its critical points and endpoints.

The Function $g(x)$

The given function is $g(x) = x \sqrt{3-x}$ . This function is defined for $x$ values between 0 and 3, inclusive. The square root function is only defined for non-negative values, so the expression inside the square root must be non-negative. This means that $3-x \geq 0$ , which implies that $x \leq 3$ . Therefore, the domain of the function is $[0, 3]$ .

Critical Points

To find the critical points of the function, we need to find the values of $x$ where the derivative of the function is zero or undefined. The derivative of the function can be found using the product rule and the chain rule.

Let's find the derivative of the function:

g'(x) = \frac{d}{dx} (x \sqrt{3-x})

Using the product rule, we get:

g'(x) = \sqrt{3-x} + x \frac{d}{dx} \sqrt{3-x}

Using the chain rule, we get:

g'(x) = \sqrt{3-x} + x \frac{1}{2\sqrt{3-x}} \frac{d}{dx} (3-x)

Simplifying the expression, we get:

g'(x) = \sqrt{3-x} + \frac{x}{2\sqrt{3-x}} (-1)

Combining the terms, we get:

g'(x) = \sqrt{3-x} - \frac{x}{2\sqrt{3-x}}

Now, we need to find the values of $x$ where the derivative is zero or undefined. Setting the derivative equal to zero, we get:

\sqrt{3-x} - \frac{x}{2\sqrt{3-x}} = 0

Multiplying both sides by $2\sqrt{3-x}$ , we get:

2(3-x) - x = 0

Simplifying the expression, we get:

6 - 2x - x = 0

Combine like terms:

6 - 3x = 0

Add $3x$ to both sides:

6 = 3x

Divide both sides by 3:

2 = x

Therefore, the critical point of the function is $x = 2$ .

Endpoints

In addition to critical points, we also need to consider the endpoints of the function. The function is defined for $x$ values between 0 and 3, inclusive. Therefore, the endpoints of the function are $x = 0$ and $x = 3$ .

Conclusion

In conclusion, the function $g(x) = x \sqrt{3-x}$ has a critical point at $x = 2$ and endpoints at $x = 0$ and $x = 3$ . The critical point indicates the presence of a maximum or minimum on the graph of the function, while the endpoints indicate the limits of the function.

Graph of the Function

The graph of the function can be visualized using a graphing calculator or software. The graph shows that the function has a maximum at $x = 2$ and decreases as $x$ approaches 0 or 3.

Mathematical Analysis

Mathematically, we can analyze the function using calculus. The derivative of the function can be used to determine the critical points and endpoints of the function. The second derivative can be used to determine the nature of the critical points.

Second Derivative

The second derivative of the function can be found by differentiating the first derivative:

g''(x) = \frac{d}{dx} \left( \sqrt{3-x} - \frac{x}{2\sqrt{3-x}} \right)

Using the chain rule and the product rule, we get:

g''(x) = \frac{1}{2\sqrt{3-x}} - \frac{1}{2\sqrt{3-x}} + \frac{x}{4(3-x)^{3/2}}

Simplifying the expression, we get:

g''(x) = \frac{x}{4(3-x)^{3/2}}

Now, we can use the second derivative to determine the nature of the critical points. If the second derivative is positive at a critical point, then the critical point is a minimum. If the second derivative is negative at a critical point, then the critical point is a maximum.

Nature of the Critical Points

Evaluating the second derivative at the critical point $x = 2$ , we get:

g''(2) = \frac{2}{4(3-2)^{3/2}} = \frac{2}{4} = \frac{1}{2}

Since the second derivative is positive at the critical point, we conclude that the critical point is a minimum.

Conclusion

Q: What is a critical point of a function?

A: A critical point of a function is a point where the derivative of the function is zero or undefined. These points are significant because they can indicate the presence of a maximum, minimum, or saddle point on the graph of the function.

Q: How do you find the critical points of a function?

A: To find the critical points of a function, you need to find the values of $x$ where the derivative of the function is zero or undefined. This can be done by setting the derivative equal to zero and solving for $x$ .

Q: What is the difference between a critical point and an endpoint?

A: A critical point is a point where the derivative of the function is zero or undefined, while an endpoint is a point where the function is defined but the derivative is not. Endpoints are typically found at the limits of the function's domain.

Q: How do you determine the nature of a critical point?

A: To determine the nature of a critical point, you need to evaluate the second derivative of the function at the critical point. If the second derivative is positive, then the critical point is a minimum. If the second derivative is negative, then the critical point is a maximum.

Q: What is the significance of critical points and endpoints in calculus?

A: Critical points and endpoints are significant in calculus because they can indicate the presence of a maximum, minimum, or saddle point on the graph of a function. This is important in optimization problems, where the goal is to find the maximum or minimum value of a function.

Q: Can you give an example of a function with a critical point and an endpoint?

A: Yes, consider the function $g(x) = x \sqrt{3-x}$ . This function has a critical point at $x = 2$ and an endpoint at $x = 3$ .

Q: How do you graph a function with critical points and endpoints?

A: To graph a function with critical points and endpoints, you can use a graphing calculator or software. You can also use the first and second derivatives of the function to determine the nature of the critical points and endpoints.

Q: What is the relationship between critical points and the second derivative?

A: The second derivative of a function can be used to determine the nature of the critical points. If the second derivative is positive at a critical point, then the critical point is a minimum. If the second derivative is negative at a critical point, then the critical point is a maximum.

Q: Can you give an example of a function with a saddle point?

A: Yes, consider the function $f(x) = x^2 - 4x + 4$ . This function has a saddle point at $x = 2$ .

Q: How do you determine the nature of a saddle point?

A: To determine the nature of a saddle point, you need to evaluate the second derivative of the function at the saddle point. If the second derivative is negative, then the saddle point is a saddle point.

Q: What is the significance of saddle points in calculus?

A: Saddle points are significant in calculus because they can indicate the presence of a local maximum or minimum on the graph of a function. This is important in optimization problems, where the goal is to find the maximum or minimum value of a function.

Conclusion

In conclusion, critical points and endpoints are significant in calculus because they can indicate the presence of a maximum, minimum, or saddle point on the graph of a function. The second derivative of a function can be used to determine the nature of the critical points and endpoints.