/*! 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 48 Express \(f\) as a composition o... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Express \(f\) as a composition of two functions; that is, find \(g\) and \(h\) such that \(f=g \circ h .\) [Note: Each exercise has more than one solution. \(]\) (a) \(f(x)=x^{2}+1\) (b) \(f(x)=\frac{1}{x-3}\)

Short Answer

Expert verified
(a) \( g(u) = u+1 \), \( h(x) = x^2 \). (b) \( g(u) = \frac{1}{u} \), \( h(x) = x-3 \).

Step by step solution

01

Identify the inner function for (a)

For the function \( f(x) = x^2 + 1 \), we need to identify an inner function \( h(x) \) to form a composition. Let \( h(x) = x^2 \), which squares the input value \( x \).
02

Identify the outer function for (a)

If \( h(x) = x^2 \), then the output of this function should be used as the input for the outer function. Let \( g(u) = u + 1 \), where \( u = h(x) = x^2 \). This means \( g(x^2) = x^2 + 1 \).
03

Express (a) as a composition

Now express \( f \) as \( f(x) = g(h(x)) \). Therefore, \( f(x) = g(h(x)) = g(x^2) = x^2 + 1 \). Thus, \( g(u) = u + 1 \) and \( h(x) = x^2 \).
04

Identify the inner function for (b)

For the function \( f(x) = \frac{1}{x-3} \), we can choose an inner function that simplifies the expression. Let \( h(x) = x - 3 \), which shifts \( x \) by 3 units.
05

Identify the outer function for (b)

If \( h(x) = x - 3 \), then the output should be used in a function that finds the reciprocal. Let \( g(u) = \frac{1}{u} \), where \( u = h(x) = x - 3 \), transforming \( u \) into its reciprocal.
06

Express (b) as a composition

Now express \( f \) as \( f(x) = g(h(x)) \). Thus, \( f(x) = g(h(x)) = g(x - 3) = \frac{1}{x-3} \). So, \( g(u) = \frac{1}{u} \) and \( h(x) = x - 3 \).

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.

Inner Function
When dealing with function composition, the "Inner Function" is a crucial component. This function typically forms the initial part of our composition. It's like laying a foundation.
Your function, let's call it "h", takes the initial input and performs the first operation on it. For example, consider the function composition in part (a) of the given exercise. We have the function \[f(x) = x^2 + 1\]. The inner function here can be chosen as \[h(x) = x^2\], which takes any input value \(x\) and squares it:
  • This operation is the first step in processing the input.
  • The output of this function, \(x^2\), serves as the input for the next function in the sequence.
So, when building your composed function, identifying the inner function is your starting point, and from there, you can design the outer operation that completes your transformation.
Outer Function
In the world of function composition, once you have identified your inner function, the next step is designing the "Outer Function". This function picks up where the inner function leaves off. Think of it as the finishing touch to our function composition process.
In the exercise given, the outer function is applied to the output of the inner function. For part (a), once we have our input squared by the function \(h(x) = x^2\),
  • We then use the outer function \(g(u) = u + 1\).
  • Here, \(u\) is the result from the inner function, \(x^2\) in this case.
  • The outer function simply adds 1 to this result, completing the operation of \(f(x)\).
Similarly, in part (b), after shifting \(x\) by 3 units with \(h(x) = x-3\),
  • The outer function \(g(u) = \frac{1}{u}\) takes over and computes the reciprocal of the outcome, \(x-3\).
  • This finalizes our composition in a complete functional transformation of the input.
The outer function is what turns the sequence of operations into the desired result.
Reciprocal Function
A "Reciprocal Function" is a special kind of function that involves flipping a number over 1 (or in simpler terms, dividing 1 by the number).
This transformation is often used to invert values, particularly useful in applications like the exercise's part (b).To express a function such as \(f(x) = \frac{1}{x-3}\) in terms of composition, we utilized a reciprocal step in our outer function.
  • Start by taking an input \(x\) and performing the shift operation through the inner function \(h(x) = x - 3\).
  • Upon producing the output \(x - 3\), we apply the reciprocal operation through the outer function \(g(u) = \frac{1}{u}\).
This step effectively transforms \(x-3\) into its reciprocal, \(\frac{1}{x-3}\).
Using reciprocal functions is a common and powerful tool in mathematical compositions for functions that need an inverse operation component like division.

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

A variable \(y\) is said to be inversely proportional to the square of a variable \(x\) if \(y\) is related to \(x\) by an equation of the form \(y=k / x^{2},\) where \(k\) is a nonzero constant, called the constant of proportionality. According to Coulomb's law, the force \(F\) of attraction between positive and negative point charges is inversely proportional to the square of the distance \(x\) between them. (a) Assuming that the force of attraction between two point charges is 0.0005 newton when the distance between them is 0.3 meter. find the constant of proportionality (with proper units). (b) Find the force of attraction between the point charges when they are 3 meters apart. (c) Make a graph of force versus distance for the two charges. (d) What happens to the force as the particles get closer and closer together? What happens as they get farther and farther apart?

To the extent that water can be assumed to be incompressible, the pressure \(p\) in a body of water varies linearly with the distance \(h\) below the surface. (a) Given that the pressure is 1 atmosphere \((1\) atm ) at the surface and 5.9 atm at a depth of \(50 \mathrm{m},\) find an equation that relates pressure to depth. (b) At what depth is the pressure twice that at the surface?

A manufacturer of cardboard drink containers wants to construct a closed rectangular container that has a square base and will hold \(\frac{1}{10}\) liter \(\left(100 \mathrm{cm}^{3}\right) .\) Estimate the dimension of the container that will require the least amount of material for its manufacture.

Use a calculating utility set to the radian mode to confirm the approximations \(\sin (\pi / 5) \approx 0.588\) and \(\cos (\pi / 8) \approx 0.924,\) and then use these values to approximate the given expressions by hand calculation. Check your answers using the trigonometric function operations of your calculating utility. (a) \(\sin \frac{4 \pi}{5}\) (b) \(\cos \left(-\frac{\pi}{8}\right)\) (c) \(\sin \frac{11 \pi}{5}\) (d) \(\cos \frac{7 \pi}{8}\) (e) \(\sin \frac{2 \pi}{5}\) (f) \(\cos ^{2} \frac{\pi}{5}\)

Parametric curves can be defined piecewise by using different formulas for different values of the parameter. Sketch the curve that is represented piecewise by the parametric equations $$\begin{aligned}&x=2 t, \quad y=4 t^{2} \quad\left(0 \leq t \leq \frac{1}{2}\right)\\\ &x=2-2 t, \quad y=2 t \quad\left(\frac{1}{2} \leq t \leq 1\right)\end{aligned}$$

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.