Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 4: Problem 56
Find formulas for the functions described. A function of the form \(y=a e^{-x}+b x\) with the global minimum at (1,2).
Short Answer
Expert verified
The function is \( y = e^{1-x} + x \).
Step by step solution
01
Set up the function
The function is given as \( y = a e^{-x} + b x \). We need to find the constants \( a \) and \( b \) such that it has a global minimum at the point \( (1, 2) \). This means that when \( x = 1 \), \( y = 2 \) and the first derivative \( y' \) at \( x = 1 \) should be zero as it is a minimum point.
02
Compute the first derivative
To find the minimum, we calculate the first derivative of the function: \( y' = \frac{d}{dx}(a e^{-x} + b x) = -a e^{-x} + b \). Set this derivative equal to zero to find the critical points: \(-a e^{-1} + b = 0 \).
03
Solve for a and b using the minimum conditions
Use the condition \( -a e^{-1} + b = 0 \) to express \( b \) in terms of \( a \): \( b = a e^{-1} \). Next, use the condition \( y(1) = 2 \) to solve for \( a \) and \( b \): \( a e^{-1} + b \times 1 = 2 \). Substitute \( b = a e^{-1} \) into the equation: \( a e^{-1} + a e^{-1} = 2 \) which simplifies to \( 2a e^{-1} = 2 \).
04
Find the value of a
Solve for \( a \) in the equation \( 2a e^{-1} = 2 \): Multiply both sides by \( e \): \( 2a = 2e \) Divide both sides by 2: \( a = e \).
05
Find the value of b
Substitute \( a = e \) into \( b = a e^{-1} \): \( b = e e^{-1} = 1 \).
06
Formulate the final function
Having found \( a = e \) and \( b = 1 \), substitute these back into the original function expression: \( y = e e^{-x} + x \). Thus, the function is \( y = e^{1-x} + x \).
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.
Function Minimum
To determine where a function reaches its lowest point, we look for the *function minimum*. This is crucial in calculus as it helps us understand the behavior of the function. For the function given by \( y = a e^{-x} + b x \), the minimum point occurs at the global minimum (1,2).
- At this minimum, the function value \( y \) at \( x = 1 \) should be 2, matching precisely the stipulated minimum coordinates.
- The reasons behind using these specific values are to ascertain where the function stabilizes at its smallest output鈥攈ence, the global minimum.
Derivatives
In calculus, *derivatives* are fundamental as they measure how a function changes as its input changes, which is crucial for finding the minimum point. For the function provided, \( y = a e^{-x} + b x \), derivative calculation helps determine how the function increases or decreases for infinitesimally small changes in \( x \).
- To find the derivative, we use the rule for differentiating exponential and linear functions: \( y' = -a e^{-x} + b \).
- The derivative tells us about the slope of the tangent line to the function at any point \( x \).
Critical Points
In calculus, *critical points* are the points on the graph of a function where the derivative is zero or undefined. They are significant because they are potential locations of local maxims and minims of the function.
- By solving the equation \( -a e^{-1} + b = 0 \), derived from setting the first derivative to zero, we identify critical points that might hold a minimum.
- The condition \( y(1) = 2 \) also verified at these critical points ensures the global minimum actually occurs at \( (1,2) \).
