Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 10: Problem 76
Find the relative maxima and relative minima, if any, of each function. $$ f(x)=x^{2} \ln x $$
Short Answer
Expert verified
Relative minimum: \(x = e^{-\frac{1}{2}}\)
No relative maxima.
Step by step solution
01
Find the first derivative
To find the first derivative of the function $$f(x) = x^{2}\ln{x}$$, we'll use the product rule, which states that if we have a function $$u(x)v(x)$$, then its derivative is:
$$
(uv)' = u'v + uv'
$$
In our case, $$u(x) = x^{2}$$ and $$v(x) = \ln{x}$$. So we first find the derivatives of each:
$$
u'(x) = 2x
\hspace{1cm}
v'(x) = \frac{1}{x}
$$
Now, we can apply the product rule to get the first derivative of our function:
$$
f'(x) = u'(x)v(x) + u(x)v'(x) = 2x\ln{x} + x^{2}\frac{1}{x} = 2x\ln{x} + x
$$
02
Determine critical points
Critical points occur where the first derivative is either zero or doesn't exist. In this case, $$f'(x) = 2x\ln{x} + x$$ is defined for all $$x > 0$$. We will solve for $$x$$ when the derivative is equal to zero:
$$
2x\ln{x} + x = 0
$$
Factor out x:
$$
x(2\ln{x} + 1) = 0
$$
There are two critical points:
1. $$x = 0$$, but this is not in the domain of the function, so we discard it.
2. $$2\ln{x} + 1 = 0$$, which leads to the critical point $$x = e^{-\frac{1}{2}}$$.
03
Use the second derivative test to classify critical points
Now, we'll find the second derivative of the function. We have $$f'(x) = 2x\ln{x} + x$$ and we'll differentiate it again to find $$f''(x)$$:
$$
f''(x) = \frac{d(2x\ln{x} + x)}{dx} = 2\ln{x} + 2 + 1 = 2\ln{x} + 3
$$
Evaluate the second derivative at the critical point $$x = e^{-\frac{1}{2}}$$:
$$
f''(e^{-\frac{1}{2}}) = 2\ln{e^{-\frac{1}{2}}} + 3 = 2(-\frac{1}{2}) + 3 = 2
$$
Since the second derivative is positive at this point, the function is concave up and the point is a relative minimum.
04
Determine relative maxima and minima
From our analysis, we have found one relative minimum at the critical point $$x = e^{-\frac{1}{2}}$$. The function has no relative maxima. So the final answer is:
- Relative minimum: $$x = e^{-\frac{1}{2}}$$
- No relative maxima.
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91影视!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Relative Maxima and Minima
When analyzing a function, one important task is to identify where it reaches its highest or lowest points in a certain interval, known as relative maxima and minima. These are the peaks and troughs in the function's graph.
Relative Maximum: This occurs at a point if the function has higher values nearby on its domain. Imagine standing on a hill鈥攜ou're at the highest point around.
Relative Minimum: This happens if the function has lower values around it, much like standing in a valley's bottom.
To detect these, we need to analyze changes in the slope, which can be done using derivatives. In the given exercise, locating the relative minima is achieved with calculus techniques that involve derivatives to understand the function's behavior and structure.
First Derivative
Derivatives are essential tools in calculus used to determine the rate at which a function is changing at any point in its domain. The first derivative represents the slope of the tangent line to the curve of the function. Here's how it plays a role in optimization:
- Finding the Derivative: For a given function, the first derivative identifies points on the graph where the slope is zero or undefined. These points are crucial because they potentially indicate where a function hits a relative maximum or minimum.
- Product Rule: In the provided problem, using the product rule helps us differentiate the product of functions like $$x^2$$ and $$\ln{x}$$. Each piece is differentiated separately and then combined.
- Solving the Derivative: By setting the first derivative equal to zero, we find critical points that might correspond to maximum or minimum values in the function.
Second Derivative Test
Once critical points are identified, the next step is to confirm whether these points are maxima, minima, or neither. The second derivative test is a powerful method to do this by analyzing the concavity of the function.
- Second Derivative Meaning: The second derivative indicates the rate of change of the first derivative, essentially telling us if the slope itself is increasing or decreasing. This is linked to the concavity of the graph of the function.
- Test Application: By calculating the second derivative at the critical points, the nature of these points is revealed. If the second derivative is positive at a point, the function is concave upwards, indicating a relative minimum. Conversely, if negative, the function is concave downwards, suggesting a relative maximum.
- Step in Solution: In the given exercise, after finding a critical point, the second derivative is positive, confirming a relative minimum at the point $$x = e^{-\frac{1}{2}}$$.
Critical Points
Critical points are the places in a function where its derivative is zero or undefined. These points are essential in calculus optimization problems as they are potential locations for relative maxima and minima. Here's what you need to know:
- Definition: A critical point occurs where the first derivative of a function is zero or does not exist. These points are the candidates for being peaks or valleys on the curve.
- Finding Them: Solving the equation given by setting the first derivative to zero often yields these points. In the discussed problem, the solution of $$f'(x) = 0$$ leads to critical points such as $$x = e^{-\frac{1}{2}}$$.
- Evaluating Validity: Not all critical points are necessarily maxima or minima. Therefore, they must be further analyzed, usually via the second derivative test, to determine their exact nature.
