Chapter 4: Problem 29
For any constant \(a\), let \(f(x)=a x-x \ln x\) for \(x>0.\) (a) What is the \(x\) -intercept of the graph of \(f(x) ?\) (b) Graph \(f(x)\) for \(a=-1\) and \(a=1.\) (c) For what values of \(a\) does \(f(x)\) have a critical point for \(x>0 ?\) Find the coordinates of the critical point and decide if it is a local maximum, a local minimum, or neither.
Short Answer
Step by step solution
Find the x-intercept
Graph for a=-1 and a=1
Find the derivative of f(x)
Find y-coordinate of critical point
Determine the nature of the critical point
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 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Derivatives
For the function \( f(x) = ax - x \ln x \), finding the derivative \( f'(x) \) helps us determine where the function has critical points. A critical point occurs when \( f'(x) = 0 \) or when the derivative is undefined. This is crucial since it indicates potential local maxima, minima, or saddle points.
For our specific function, the derivative is \( f'(x) = a - \ln x - 1 \). Setting \( f'(x) = 0 \) simplifies to \( \ln x = a - 1 \), leading to \( x = e^{a-1} \). This result gives us clues about the behavior of \( f(x) \) at this specific \( x \)-value.
Local Maximum
To determine whether a critical point is a local maximum, we use the second derivative test. For our function, once we obtain the critical point \( (e^{a-1}, f(e^{a-1})) \), we calculate the second derivative \( f''(x) = -\frac{1}{x} \).
Substituting \( x = e^{a-1} \) into the second derivative gives \( f''(e^{a-1}) = -1 / e^{a-1} \), which is negative. This negativity indicates that the function is concave down at this point, confirming it as a local maximum.
- Negative second derivative: Concave down, hence a local maximum.
Graphing Functions
The graph for \( a = -1 \) would be \( f(x) = -x - x \ln x \). Here, the 'weight' of \( x \) is negative, influencing the curve to shift downwards. Conversely, for \( a = 1 \), the function is \( f(x) = x - x \ln x \), presenting a different slope and curvature, affecting how steep or gentle the graph appears over its domain.
Graphing Tips:
- Identify key features: x-intercepts, turning points, and asymptotic behavior.
- Note changes with parameters: As \( a \) changes, observe shifts and modifications in graph shape.
- Understand basic curve families: Knowing how parent curves behave helps graph transformations.
Calculus
In our example, we utilize calculus to achieve several tasks: find derivatives, identify critical points, and analyze the function's behavior. By applying methods such as the product rule and logarithmic differentiation, we can manipulate complex expressions like \( ax - x \ln x \).
Key Takeaways from Calculus in this Exercise:
- Derivatives: Capture the rate of change, helping to find where the function shifts from increasing to decreasing.
- Critical Points: Determine where peaks and troughs of the function occur, important for understanding maximum and minimum values.
- Graph Analysis: Application of derivatives reveals a lot about the overall shape and important characteristics of functions.