/*! 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 For each pair of functions, find... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

For each pair of functions, find \(f(g(x))\) and \(g(f(x)) .\) Simplify your answers. $$ f(x)=\sqrt[3]{x}, g(x)=\frac{x+1}{x^{3}} $$

Short Answer

Expert verified
\( f(g(x)) = \frac{(x+1)^{\frac{1}{3}}}{x} \), \( g(f(x)) = \frac{\sqrt[3]{x} + 1}{x} \).

Step by step solution

01

Understanding the Problem

We have two functions, \( f(x) = \sqrt[3]{x} \) and \( g(x) = \frac{x+1}{x^{3}} \). We are tasked with finding and simplifying the compositions \( f(g(x)) \) and \( g(f(x)) \).
02

Find \( f(g(x)) \) Composition

To find \( f(g(x)) \), we substitute \( g(x) = \frac{x+1}{x^{3}} \) into the function \( f \). This gives us: \[ f(g(x)) = f\left(\frac{x+1}{x^{3}}\right) = \sqrt[3]{\frac{x+1}{x^{3}}} \].
03

Simplify \( f(g(x)) \)

We simplify \( \sqrt[3]{\frac{x+1}{x^{3}}} \) by expressing it as \[ \frac{(x+1)^{\frac{1}{3}}}{(x^{3})^{\frac{1}{3}}} = \frac{(x+1)^{\frac{1}{3}}}{x} \]. Thus, \( f(g(x)) = \frac{(x+1)^{\frac{1}{3}}}{x} \).
04

Find \( g(f(x)) \) Composition

Next, we find \( g(f(x)) \) by substituting \( f(x) = \sqrt[3]{x} \) into the function \( g \). Thus, \[ g(f(x)) = g(\sqrt[3]{x}) = \frac{\sqrt[3]{x} + 1}{(\sqrt[3]{x})^{3}} \].
05

Simplify \( g(f(x)) \)

The simplification of \( g(f(x)) \) simplifies \( (\sqrt[3]{x})^{3} = x \), giving us: \[ \frac{\sqrt[3]{x} + 1}{x} \]. Thus, \( g(f(x)) = \frac{\sqrt[3]{x} + 1}{x} \).

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 91Ó°ÊÓ!

Key Concepts

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

Inverse Functions
Inverse functions are special types of functions that reverse the effect of the original function. If you have a function say, \( f(x) \), then its inverse \( f^{-1}(x) \) will "undo" whatever \( f(x) \) does. For example, if \( f(x) \) adds 3 to \(x\), \( f^{-1}(x) \) will subtract 3 to bring \(x\) back to its original value.
When dealing with inverse functions, it's like having a reversible process.
To check if two functions are inverses, you should perform function composition both ways, \( f(g(x)) \) and \( g(f(x)) \). Both compositions should give you back \( x \) if the functions are indeed inverses. In our exercise, we examined two functions, but no indication was given that they are inverses.
Inverse functions are common in many areas of math, providing powerful methods to solve equations and translate problems from one form into another, more solvable form.
Function Notation
Function notation is a way of representing functions in mathematics that declares a relationship between input and output values. When you see something like \( f(x) \), the letter \( f \) signifies the name of the function, and \( x \) is the variable or input.
The beauty of function notation is its clarity and simplicity.
It allows mathematicians and students to easily understand which variable is being manipulated and what operations are involved.

In our problem, we used function notation to represent two distinct functions: \( f(x) = \sqrt[3]{x} \), which is the cube root function, and \( g(x) = \frac{x+1}{x^{3}} \), a rational function.
Using these notations, we can easily substitute one function into another, as shown in the solution, to explore combinations and interactions such as \( f(g(x)) \) and \( g(f(x)) \).
Always use function notation to keep your work organized and avoid confusion, especially with complex functions or equations.
Simplification of Expressions
Simplifying expressions is a critical skill in mathematics, where you aim to reduce expressions to their simplest or most manageable form. This often involves combining like terms, factoring, or performing arithmetic operations to make an expression easier to work with or solve.

In the exercise, we simplified expressions resulting from function composition. After substituting \( g(x) \) into \( f(x) \) to find \( f(g(x)) \), we simplified \( \sqrt[3]{\frac{x+1}{x^{3}}} \) into \( \frac{(x+1)^{\frac{1}{3}}}{x} \). Similarly, for \( g(f(x)) \), we simplified \( \frac{\sqrt[3]{x} + 1}{x} \).
The simplification mainly involved applying the laws of exponents, such as \( (a^m)^n = a^{m\times n} \).

Ultimately, simplification is about reducing clutter in expressions, thus allowing easier manipulation, solving, or integration of these expressions in further calculations.

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

Starting with the graph of \(f(x)=6^{x}\) write the equation of the graph that results from a. reflecting \(f(x)\) about the \(x\) -axis and the \(y\) -axis b. reflecting \(f(x)\) about the \(x\) -axis, shifting left 2 units, and down 3 units

Describe how each function is a transformation of the original function \(f(x)\). $$ f(-x) $$

Select all of the following tables which represent \(y\) as a function of \(x\) and are one-to-one. a. $$ \begin{array}{|l|l|l|l|} \hline x & 3 & 8 & 12 \\ \hline y & 4 & 7 & 7 \\ \hline \end{array} $$ b. $$ \begin{array}{|l|l|l|l|} \hline x & 3 & 8 & 12 \\ \hline y & 4 & 7 & 13 \\ \hline \end{array} $$ c. $$ \begin{array}{|l|l|l|l|} \hline x & 3 & 8 & 12 \\ \hline y & 4 & 7 & 7 \\ \hline \end{array} $$

Dave leaves his office in Padelford Hall on his way to teach in Gould Hall. Below are several different scenarios. In each case, sketch a plausible (reasonable) graph of the function \(s=d(t)\) which keeps track of Dave's distance \(s\) from Padelford Hall at time \(t\). Take distance units to be "feet" and time units to be "minutes." Assume Dave's path to Gould Hall is long a straight line which is 2400 feet long. [UW] a. Dave leaves Padelford Hall and walks at a constant spend until he reaches Gould Hall 10 minutes later. b. Dave leaves Padelford Hall and walks at a constant speed. It takes him 6 minutes to reach the half-way point. Then he gets confused and stops for 1 minute. He then continues on to Gould Hall at the same constant speed he had when he originally left Padelford Hall. c. Dave leaves Padelford Hall and walks at a constant speed. It takes him 6 minutes to reach the half-way point. Then he gets confused and stops for 1 minute to figure out where he is. Dave then continues on to Gould Hall at twice the constant speed he had when he originally left Padelford Hall. d. Dave leaves Padelford Hall and walks at a constant speed. It takes him 6 minutes to reach the half-way point. Then he gets confused and stops for 1 minute to figure out where he is. Dave is totally lost, so he simply heads back to his office, walking the same constant speed he had when he originally left Padelford Hall. e. Dave leaves Padelford heading for Gould Hall at the same instant Angela leaves Gould Hall heading for Padelford Hall. Both walk at a constant speed, but Angela walks twice as fast as Dave. Indicate a plot of "distance from Padelford" vs. "time" for the both Angela and Dave. f. Suppose you want to sketch the graph of a new function \(\mathrm{s}=\mathrm{g}(\mathrm{t})\) that keeps track of Dave's distance s from Gould Hall at time t. How would your graphs change in (a)-(e)?

Suppose \(f(x)=x^{2}+8 x-4\). Compute the following: $$ \text { a. } f(-1)+f(1) \quad \text { b. } f(-1)-f(1) $$

See all solutions

Recommended explanations on Math 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.