Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 4: Problem 57
Find formulas for the functions described. A function of the form \(y=b x e^{-a x}\) with a local maximum at (3,6).
Short Answer
Expert verified
The function is \( y = 2x \cdot e^{1-\frac{x}{3}}. \)
Step by step solution
01
Understand the function form
The function we are given is of the form \( y = b x e^{-a x} \), which combines polynomial and exponential components. Our task is to determine the constants \(a\) and \(b\) such that the function has a local maximum at the point \((3,6)\).
02
Differentiate the function
To find where the local maximum occurs, differentiate the function. We use the product rule for differentiation:\[ \frac{d}{dx}(b x e^{-a x}) = b \left(1 \cdot e^{-a x} + x \cdot \frac{d}{dx}(e^{-a x}) \right) = b \left(e^{-a x} - a x e^{-a x} \right) = b e^{-a x}(1 - a x). \]
03
Set the derivative to zero for critical points
Set the derivative \( b e^{-a x}(1 - a x) = 0 \) to find the critical points. Since \( b e^{-a x} eq 0 \), the equation simplifies to \[ 1 - a x = 0 \] which gives \[ a x = 1 \] or \[ x = \frac{1}{a}. \]
04
Use the given local maximum point
Since the function has a local maximum at \(x = 3\), we set \[ \frac{1}{a} = 3 \] which gives \[ a = \frac{1}{3}. \]
05
Substitute the maximum point into the original function
We know the function value is 6 at \(x = 3\):\[ y = b \cdot 3 \cdot e^{-\frac{1}{3} \cdot 3} = 6. \] This simplifies to \[ 3b \cdot e^{-1} = 6. \] Solving for \(b\), we have \[ b = \frac{6}{3} \cdot e = 2e. \]
06
Write the final function
Substitute \(a = \frac{1}{3}\) and \(b = 2e\) into the original function form to get the final formula:\[ y = 2e \cdot x \cdot e^{-\frac{1}{3}x}. \] This simplifies to:\[ y = 2x \cdot e^{1-\frac{1}{3}x}. \]
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.
Differentiation
Differentiation is a cornerstone of calculus that involves finding the derivative of a function. Knowing how to differentiate properly can lead us to important insights such as rates of change, tangents to curves, and optimizing problems.
- The derivative of a function at a particular point on its curve tells us the slope of the tangent line at that point, which is essentially how fast the function is changing at that very spot.
- In the context of this exercise, we differentiate the function \( y = b x e^{-a x} \) using the product rule. The product rule is used here because \( y \) is a product of two separate functions \( b x \) and \( e^{-a x} \).
- Applying the product rule involves differentiating each component and adding the results: \( \ \frac{d}{dx}(b x e^{-a x}) = b(e^{-a x} - a x e^{-a x}) \).
Critical Points
Critical points on a function are values of \( x \) where the derivative is zero or undefined. These points are particularly significant because they indicate potential maxima, minima, or saddle points.
- In our example, after differentiating the function \( y = b x e^{-a x} \), the resulting derivative \( b e^{-a x}(1 - a x) \) leads us to critical points when set to zero.
- Critical points occur at \( x = \frac{1}{a} \), found by solving \( 1 - a x = 0 \).
Local Maximum
A function's local maximum is a point where the function value is greater than all nearby values. This is crucial for understanding the behavior of the function around specific values.
- To determine a local maximum, one locates the critical points and evaluates the function's behavior around these points, often with tests such as the first or second derivative test.
- In this problem, the statement of a local maximum at \( x=3 \) simplifies to calculating \( a \) when \( 1 - a x = 0 \), giving \( a = \frac{1}{3} \).
- It's confirmed by substituting \( x = 3 \) into the function, ensuring the function peaks at 6 at this point.
