/*! 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 26 Find a formula for the inverse o... [FREE SOLUTION] | 91Ó°ÊÓ

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Find a formula for the inverse of the function. \(y=\frac{e^{x}}{1+2 e^{x}}\)

Short Answer

Expert verified
The inverse function is \(y = \ln\left(\frac{x}{1 - 2x}\right)\).

Step by step solution

01

Understand the Problem

You are given a function \(y = \frac{e^{x}}{1 + 2e^{x}}\). The goal is to find a formula for its inverse. Inverse functions reverse the roles of inputs and outputs, so we want to express \(x\) in terms of \(y\).
02

Swap Variables

To find the inverse, replace \(y\) with \(x\) and \(x\) with \(y\). This gives us the equation: \(x = \frac{e^{y}}{1 + 2e^{y}}\). Now, solve this equation for \(y\).
03

Solve for the Exponential Term

Multiply both sides by \(1 + 2e^{y}\) to eliminate the fraction: \(x (1 + 2e^{y}) = e^{y}\). Simplify to get: \(x + 2xe^{y} = e^{y}\).
04

Rearrange the Equation

Rearrange the equation to isolate terms involving \(e^{y}\): \(e^{y} - 2xe^{y} = x\). Factor \(e^{y}\) to get: \(e^{y}(1 - 2x) = x\).
05

Solve for \(e^{y}\)

Divide both sides by \(1 - 2x\) to solve for \(e^{y}\): \(e^{y} = \frac{x}{1 - 2x}\).
06

Take the Natural Logarithm

Take the natural logarithm of both sides to solve for \(y\): \(y = \ln\left(\frac{x}{1 - 2x}\right)\).
07

Confirm the Inverse Relationship

To ensure correctness, check that substituting \(x = \ln\left(\frac{y}{1 - 2y}\right)\) back into the original function yields \(y\) as expected.

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Key Concepts

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

Exponential Functions
Exponential functions are a special type of mathematical function where the variable appears in the exponent. In our exercise, the function given is \( y = \frac{e^x}{1 + 2e^x} \), which includes the natural exponential function \( e^x \). Here, \( e \) is the base of the natural logarithm, approximately equal to 2.71828. Exponential functions grow rapidly and are widely used in various disciplines like finance, physics, and biology for modeling growth processes.
  • Exponential Growth: As \( x \) increases, \( e^x \) grows exponentially, meaning very fast.
  • Asymptotic Behavior: Exponential functions often approach certain values, known as asymptotes, but never actually reach them.
Understanding how to manipulate these functions is crucial for solving problems involving growth or decay, such as population models or radioactive decay, where they frequently appear.
When dealing with inverse functions involving exponentials, the natural logarithm becomes a powerful tool to "undo" the exponential part, making it easier to solve for the variable.
Natural Logarithms
Natural logarithms, denoted as \( \ln(x) \), are the inverse of the exponential function \( e^x \). In our worked example, we ultimately used natural logarithms to solve for \( y \) once we had isolated \( e^y \). The equation \( y = \ln\left(\frac{x}{1 - 2x}\right) \) was derived by applying the natural logarithm to both sides of the equation \( e^y = \frac{x}{1 - 2x} \).
  • Inverse Function: Taking the natural log of \( e^y \) returns the exponent, \( y \), because \( \ln(e^y) = y \).
  • Logarithmic Properties: \( \ln(a \cdot b) = \ln(a) + \ln(b) \) and \( \ln\left(\frac{a}{b}\right) = \ln(a) - \ln(b) \).
Natural logarithms are very helpful in solving equations where the unknown variable is in the exponent. By applying logarithmic identities and properties, you can simplify complicated expressions and solve them systematically.
Function Transformation
Function transformation involves changing the appearance of a function through operations like shifts, stretches, and reflections. In our exercise, while finding the inverse, we notably transformed the function by swapping \( x \) and \( y \), which is a key step in deriving inverses for any function.
  • Variable Swapping: To find an inverse, switch the roles of the input and output variables. It transforms the original function.
  • Solving for New Output: After swapping, you usually have to solve for the new output to finish the transformation. Here, this involved manipulating the equation to isolate \( e^y \) and later applying a logarithm.
  • Checking Work: Transformation typically ends with verification. Inverse functions should "undo" each other, meaning plugging one into the other should return the input.
Function transformations are essential tools for manipulating functions to explore their properties and solve equations. They form a bridge between a function and its inverse, helping to visualize and understand complex relationships.

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Most popular questions from this chapter

A cell phone plan has a basic charge of \(\$ 35\) a month. The plan includes 400 free minutes and charges 10 cents for each additional minute of usage. Write the monthly cost \(C\) as a function of the number \(x\) of minutes used and graph \(C\) as a function of \(x\) for 0\(\leqslant x \leqslant 600\) .

When a camera flash goes off, the batteries immediately begin to recharge the flash's capacitor, which stores electric charge given by $$Q(t)=Q_{0}\left(1-e^{-t / a}\right)$$ (The maximum charge capacity is \(Q_{0}\) and \(t\) is measured inseconds.) (a) Find the inverse of this function and explain its meaning. (b) How long does it take to recharge the capacitor to 90\(\%\) of capacity if \(a=2 ?\)

Suppose that the graph of \(y=\log _{2} x\) is drawn on a coordinate grid where the unit of measurement is an inch. How many miles to the right of the origin do we have to move before the height of the curve reaches 3 \(\mathrm{ft}\) ?

Express the function in the form \(f \circ g\) \(v(t)=\sec \left(t^{2}\right) \tan \left(t^{2}\right)\)

Bioavailability is a term that refers to the fraction of an antibiotic dose taken orally that is absorbed into the bloodstream. Suppose that, for a dosage of \(x\) mg, the bioavailability is \(h(x)=\frac{1}{2} x\) mg. If \(x\) mg enters the bloodstream, suppose that the amount eventually absorbed into the site of an infection is given by \(g(x)=4 x /(x+4)\) mg. Finally, if \(x\) mg is absorbed into the site of an infection, suppose that the number of surviving bacteria is given by \(f(x)=3200 /\left(8+x^{2}\right),\) measuredin colony forming units, CFU. (a) Derive the function that relates oral dosage to the number of surviving bacteria using composition of functions. (b) Suppose the antibiotic is instead administered by injection. Derive the function that relates dosage to the number of surviving bacteria using composition of functions. (c) Sketch the graphs of the functions found in parts (a)

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