/*! 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 78 Find the inverse of \(f(x)=\left... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Find the inverse of \(f(x)=\left(\frac{1}{3}\right)^{x}, x \in \mathbf{R}\), together with its domain, and graph both functions in the same coordinate system.

Short Answer

Expert verified
The inverse is \(f^{-1}(x) = \frac{\ln(x)}{\ln\left(\frac{1}{3}\right)}\) with domain \(x > 0\).

Step by step solution

01

Understanding the Concept of an Inverse Function

The inverse of a function "undoes" the action of the function. For a function \(f(x)\), its inverse \(f^{-1}(x)\) satisfies \(f(f^{-1}(x)) = x\) and \(f^{-1}(f(x)) = x\). Our task is to find the inverse of \(f(x) = \left(\frac{1}{3}\right)^x\).
02

Set Up the Equation for the Inverse

To find the inverse function, start by letting \(y = f(x)\), so \(y = \left(\frac{1}{3}\right)^x\). To find \(x\) in terms of \(y\), we need to solve this equation for \(x\).
03

Solve for x in Terms of y

Take the logarithm of both sides to solve for \(x\). We can choose any logarithm base, but natural logarithm (\(\ln\)) is commonly used: \[ \ln(y) = \ln\left(\left(\frac{1}{3}\right)^x\right) \]Using the power rule of logarithms (\(\ln(a^b) = b \cdot \ln(a)\)), we have:\[ \ln(y) = x \cdot \ln\left(\frac{1}{3}\right) \]Therefore, solving for \(x\):\[ x = \frac{\ln(y)}{\ln\left(\frac{1}{3}\right)} \]
04

Write the Inverse Function

Now replace \(x\) with \(f^{-1}(y)\), giving us the inverse function:\[ f^{-1}(x) = \frac{\ln(x)}{\ln\left(\frac{1}{3}\right)} \]
05

Find the Domain of the Inverse Function

The domain of the inverse function \(f^{-1}(x)\) is the range of the original function \(f(x)\). Since \(f(x) = \left(\frac{1}{3}\right)^x\) is defined for all real \(x\) and yields positive values only, the domain of \(f^{-1}(x)\) is \(x > 0\).
06

Graphing Both Functions

When graphing, plot \(f(x) = \left(\frac{1}{3}\right)^x\) and its inverse \(f^{-1}(x) = \frac{\ln(x)}{\ln\left(\frac{1}{3}\right)}\) on the same axes. The two graphs will be reflections of each other across the line \(y = x\). Ensure that on the graph of \(f^{-1}(x)\), \(x\) is always greater than 0.

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.

Exponential Function
An exponential function is a mathematical expression where a constant base is raised to a variable exponent. Specifically, in the exercise given, the function is defined as \( f(x) = \left(\frac{1}{3}\right)^x \). Here, the base is \( \frac{1}{3} \) and the exponent, \( x \), is a variable.

Key characteristics of exponential functions include:
  • Rapid growth or decay: These functions grow or decrease at increasingly faster rates as \( x \) changes.
  • Never reaching zero: The graph approaches a horizontal asymptote (usually the x-axis) but never actually touches it.
  • Continuity: Exponential functions are continuous and infinitely differentiable.
In our exercise, since the base \( \left(\frac{1}{3}\right) \) is less than 1, the function demonstrates exponential decay, meaning it decreases as \( x \) increases. By understanding this behavior, students can better grasp how these functions are applied in different real-world contexts like radioactive decay or population shrinkage.
Domain of a Function
The domain of a function refers to all possible input values (\( x \)) for which the function is defined. For the original exponential function, \( f(x) = \left(\frac{1}{3}\right)^x \), the domain is all real numbers (\( x \in \mathbf{R} \)), as you can raise \( \frac{1}{3} \) to any real number.

Understanding the domain helps in determining where the function is applicable or valid. In the case of its inverse, \( f^{-1}(x) = \frac{\ln(x)}{\ln\left(\frac{1}{3}\right)} \), the inverse is only defined for positive \( x \), thus its domain is \( x > 0 \).

This concept is vital when dealing with inverse functions since the domain of the inverse is the range of the original function, and vice versa. By determining these domains, students can ensure that they solve problems only within the valid scope of the function.
Logarithms
Logarithms are the inverse operations of exponentials. They answer the question: "To what power must a certain base be raised, to produce a given number?"

In our exercise, we used logarithms to find the inverse of the exponential function. By taking the natural logarithm (\( \ln \)) of both sides in the equation \( y = \left(\frac{1}{3}\right)^x \), we were able to isolate \( x \).

Key points about logarithms:
  • They help in solving equations involving exponentials, especially when the variable is in the exponent.
  • The logarithm of a power, \( \ln(a^b) = b \cdot \ln(a) \), is a critical rule to remember.
  • The inverse relationship between logarithms and exponentials allows conversion between forms easily.
Utilizing these properties, we rearranged to find the inverse function of \( f(x) \). Developing a firm understanding of logarithms empowers students to tackle various mathematical problems involving growth and decay models, like those seen in economics or biology.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The size distribution of zooplankton in a lake is typically a hump-shaped curve; that is, the number of zooplankton of a given size increases with size up to a critical size and then decreases with size for organisms larger than that critical size. Brooks and Dodson (1965) studied the effects of introducing a planktivorous fish in a lake. They found that the composition of zooplankton after the fish was introduced shifted to smaller individuals. In the same coordinate system, sketch the size distribution of zooplankton before and after the introduction of the planktivorous fish.

Suppose that \(N(t)\) denotes a population size at time \(t\) and satisfies the equation $$ N(t)=2 e^{3 t} \quad \text { for } t \geq 0 $$ (a) If you graph \(N(t)\) as a function of \(t\) on a semilog plot, a straight line results. Explain why. (b) Graph \(N(t)\) as a function of \(t\) on a semilog plot, and determine the slope of the resulting straight line.

sketch the graph of each function. Do not use a graphing calculator. (Assume the largest possible domain.) $$ y=2 \sin (x+\pi / 4) $$

Michaelis-Menten Equation Enzymes serve as catalysts in many chemical reactions in living systems. The simplest such reactions transform a single substrate into a product with the help of an enzyme. The Michaelis-Menten equation describes the rate of such enzymatically controlled reactions. The equation, which gives the relationship between the initial rate of the reaction, \(v\), and the concentration of the substrate, \(s\), is $$ v(s)=\frac{v_{\max } s}{s+K} $$ where \(v_{\max }\) is the maximum rate at which the product may be formed and \(K\) is called the Michaelis-Menten constant. Note that this equation has the same form as the Monod growth function. Given some data on the reaction rate \(v\), for different substrate concentrations \(s\), we would like to infer the parameters \(K\) and (a) The graph of \(v\) against \(s\) is nonlinear, so it is hard to determine \(K\) and \(v_{\max }\) directly from a graph of the function \(v(s) .\) In the remaining parts of this question you will be guided to transform your plot into one in which the dependent variable depends linearly on the independent variable. First plot, using a graphing calculator, or by hand, \(v(s)\) for the following values of \(K\) and \(v_{\max }\) : $$ \left(K, v_{\max }\right)=(1,1), \quad(2,1), \quad(1,2) $$ (b) Show that the Michaelis-Menten equation can be written in the form $$ \frac{1}{v}=\frac{K}{v_{\max }} \frac{1}{s}+\frac{1}{v_{\max }} $$

Find the amplitude and the period of \(f(x)\) $$ f(x)=-\frac{3}{2} \sin \left(\frac{\pi}{3} x\right), \quad x \in \mathbf{R} $$

See all solutions

Recommended explanations on Biology Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.