/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 41 Find formulas for the functions ... [FREE SOLUTION] | 91Ó°ÊÓ

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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 \( y= \frac{2e}{1+e^{-1}} e^{-x} + \frac{2}{1+e^{-1}} x \).

Step by step solution

01

Understanding the Function Form

Given the function of the form \( y = a e^{-x} + b x \), where \( e \) is the base of the natural logarithm, \( a \) and \( b \) are constants, and \( x \) is the independent variable.
02

Differentiating the Function

To find the minimum, we first differentiate the function: \( y = a e^{-x} + b x \). The derivative is \( y' = -a e^{-x} + b \).
03

Setting the Derivative to Zero at the Minimum

Since the minimum occurs where the derivative is zero, set \( y' = -a e^{-x} + b = 0 \). At \( x=1 \), substitute to get \( -a e^{-1} + b = 0 \), which simplifies to \( b = \frac{a}{e} \).
04

Using the Global Minimum to Find Constants

The global minimum value is \( y(1) = 2 \). Substitute \( x = 1 \) into the original function: \( 2 = a e^{-1} + b \cdot 1 \). Substitute \( b = \frac{a}{e} \) into this equation to get \( 2 = a e^{-1} + \frac{a}{e} \).
05

Solving for \(a\)

Simplifying the equation from Step 4: \( 2 = \frac{a}{e} (e^{-1} + 1) \). Multiply through by \( e \): \( 2e = a (1 + e^{-1}) \). Solve for \( a \) to find \( a = \frac{2e}{1+e^{-1}} \).
06

Solving for \(b\)

Substitute \( a \) back into \( b = \frac{a}{e} \) to find \( b = \frac{2e}{e(1+e^{-1})} = \frac{2}{1+e^{-1}} \).
07

Formula Summary

Thus, the function is \( y = \frac{2e}{1+e^{-1}} e^{-x} + \frac{2}{1+e^{-1}} x \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Differentiation
Differentiation is a core concept in calculus that allows us to find the rate at which a function is changing at any given point. When given a function, such as \( y = a e^{-x} + b x \), differentiation helps us determine the slope of the tangent line to the curve of the function at any point along its path.

In practical scenarios like the one given, we differentiate to find where the function reaches its extreme values, such as minimums or maximums. To differentiate \( y = a e^{-x} + b x \), we use the basic rules of differentiation: The derivative of \( e^{-x} \) is \( -e^{-x} \), and the derivative of \( b x \) is simply \( b \). This gives us the derivative, \( y' = -a e^{-x} + b \).
  • Derivatives reveal the behavior of functions.
  • They are essential for solving optimization problems.
Finding global minimum
Finding a global minimum involves identifying the lowest point of a function over its entire domain. This is crucial in calculus because it helps us understand the function's behavior and optimize it for specific purposes.

To find where this occurs, we set the derivative of the function to zero. This tells us where the slope of the tangent to the function is horizontal, indicating potential minimum or maximum points. For the function \( y = a e^{-x} + b x \), setting its derivative \( y' = -a e^{-x} + b \) equal to zero allows us to find the critical points.

By solving \( -a e^{-x} + b = 0 \), we can locate where exactly the global minimum is, which is specified in this problem to be at \( x=1 \).
  • Global minimum is found by solving \( y' = 0 \).
  • Calculating the value of the function at these points verifies them.
Exponential function
An exponential function possesses a characteristic rate of change that is proportional to its current value. They have the form \( y = a e^{-x} \), where \( e \) is a constant approximately equal to 2.71828. In the context of our specific function, \( e^{-x} \) decreases as \( x \) increases, which gives exponential decay.

Understanding exponential functions is vital because they appear regularly in natural and financial scenarios, such as biological decay and interest compounding. In the equation \( y = a e^{-x} + b x \), we see \( e^{-x} \) representing decay, balanced by the linear term \( b x \).
  • Exponential decay "shrinks" the function rapidly as \( x \) increases.
  • These functions show a sharp decline before leveling out.
Constants in functions
Constants in functions play a crucial role in defining the shape and position of graphs. In the function \( y = a e^{-x} + b x \), \( a \) and \( b \) are constants.

The constant \( a \) determines how steeply the exponential part of the function behaves. With a rise in \( a \), the curve becomes more pronounced. Similarly, the constant \( b \) dictates the slope of the linear component of the function. If \( b \) increases, the linear component tilts more steeply upwards or downwards, depending on its sign.
  • Constant \( a \) affects exponential growth or decay rate.
  • Constant \( b \) adjusts linear growth or decline.
Knowing these constants and how to manipulate and calculate them is essential, especially when achieving specific results like finding a global minimum, which depends on substituting specific values and solving linear equations to match given conditions like \((1, 2)\).

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Most popular questions from this chapter

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