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Chapter 2: Problem 54

Express the function in the form \(f \circ g\) $$H(x)=\sqrt{1+\sqrt{x}}$$

Short Answer

Expert verified
The function H(x) is expressed as a composite: \( f \circ g \) where \( f(u) = \sqrt{u} \) and \( g(x) = 1 + \sqrt{x} \).

Step by step solution

01

Understand Composite Functions

Composite functions are when one function is applied to the result of another function. In other words, if you have two functions \( f \) and \( g \), then the composite function \( f \circ g \) is defined as \( f(g(x)) \). Our goal is to rewrite the given function in this form.
02

Identify Outer and Inner Functions

Analyze the structure of the function \( H(x) = \sqrt{1 + \sqrt{x}} \). The outermost operation applied to \( x \) is the square root, and inside of this, there is another expression, \( 1 + \sqrt{x} \). Thus, we can separate the function into an outer function \( f(u) = \sqrt{u} \) and an inner function \( g(x) = 1 + \sqrt{x} \).
03

Define the Functions

Start by defining the inner function: \( g(x) = 1 + \sqrt{x} \). This function takes \( x \), computes its square root, and adds 1. The outer function \( f(u) = \sqrt{u} \) takes the input from \( g(x) \) and calculates the square root of the result.
04

Write the Composite Function

Express the function \( H(x) \) as a composition of \( f \) and \( g \). By the definition of composition, \( f(g(x)) = f(1 + \sqrt{x}) = \sqrt{1 + \sqrt{x}} \), which matches the given function \( H(x) \). Therefore, \( H(x) = f(g(x)) \).
05

Verify the Composition

To ensure correctness, substitute \( g(x) \) into \( f \) and check: \( f(g(x)) = f(1 + \sqrt{x}) = \sqrt{1 + \sqrt{x}} \). This is indeed \( H(x) \), confirming that the composition is correct.

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Key Concepts

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

Function Composition
Understanding function composition is vital in mathematics. It involves taking two functions and applying one to the result of the other. This simplifies complex expressions by breaking them into simpler parts. For example, if you have functions \(f(x)\) and \(g(x)\), the composite function \(f \circ g\) is expressed as \(f(g(x))\). Think of it as a machine where \(g(x)\) processes the input first, and \(f\) further processes \(g(x)\)'s output.

Let's apply this concept to solve a problem using the function \(H(x) = \sqrt{1 + \sqrt{x}}\):
  • First, recognize the two functions involved.
  • Express \(H(x)\) in terms of an outer function \(f(u)\) and an inner function \(g(x)\).
  • By doing this, you simplify the process of evaluating complex expressions.
This clear simplification is what makes function composition such a powerful tool.
Outer and Inner Functions
To decompose a complex function like \(H(x) = \sqrt{1 + \sqrt{x}}\), it's crucial to identify its outer and inner functions. The outer function performs the last operation, wrapping around the inner workings. The operation closest to \(x\) is considered the inner function.

In our exercise:
  • The inner function, \(g(x) = 1 + \sqrt{x}\), performs two tasks. It calculates the square root of \(x\) and adds 1.
  • The outer function, \(f(u) = \sqrt{u}\), simply takes \(u\) and finds its square root.
Separating functions in this way reveals how each part operates on the variable \(x\). Mastering this identification process leads to better understanding and easier manipulation of functions.
Mathematical Problem-Solving
Applying function composition is a significant aspect of mathematical problem-solving. It calls for analytical thinking and understanding how various components work together.

When faced with a problem like expressing \(H(x) = \sqrt{1 + \sqrt{x}}\) in the form \(f \circ g\):
  • Break down the problem by identifying and defining the required functions.
  • Verify the function composition by substituting the inner function back into the outer function.
  • Check your steps for accuracy by ensuring they resolve back to the original expression.
This method strengthens logical reasoning skills and sharpens critical thinking by encouraging a structured approach to problem-solving.
Precalculus Concepts
Function composition plays a crucial role in precalculus. It paves the way for understanding more advanced topics in calculus and beyond.

The current exercise exemplifies key precalculus concepts like:
  • Understanding function operations and transformations.
  • Grasping the relationship between different functions in a composed format.
  • Developing problem-solving methods to tackle complex expressions.
These foundational skills are not only necessary for calculus but also benefit numerous applications in engineering, physics, and computer science.

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Most popular questions from this chapter

Evaluate the function at the indicated values. $$\begin{aligned} &h(t)=t+\frac{1}{t} ; \\ &h(1), h(-1), h(2), h\left(\frac{1}{2}\right), h(x), h\left(\frac{1}{x}\right) \end{aligned}$$

Find the domain of the function. $$f(t)=\sqrt[3]{t}-1$$

A highway engineer wants to estimate the maximum number of cars that can safely travel a particular highway at a given speed. She assumes that each car is \(17 \mathrm{ft}\) long, travels at a speed \(s,\) and follows the car in front of it at the "safe following distance" for that speed. She finds that the number \(N\) of cars that can pass a given point per minute is modeled by the function $$N(s)=\frac{88 s}{17+17\left(\frac{s}{20}\right)^{2}}$$ At what speed can the greatest number of cars travel the highway safely?

At the beginning of this section we discussed three examples of everyday, ordinary functions: Height is a function of age, temperature is a function of date, and postage cost is a function of weight. Give three other examples of functions from everyday life.

As blood moves through a vein or an artery, its velocity \(v\) is greatest along the central axis and decreases as the distance \(r\) from the central axis increases (see the figure). The formula that gives \(v\) as a function of \(r\) is called the law of laminar flow. For an artery with radius \(0.5 \mathrm{cm},\) the relationship between \(v\) (in \(\mathrm{cm} / \mathrm{s}\) ) and \(r\) (in \(\mathrm{cm}\) ) is given by the function $$v(r)=18,500\left(0.25-r^{2}\right) \quad 0 \leq r \leq 0.5$$ (a) Find \(v(0.1)\) and \(v(0.4)\) (b) What do your answers to part (a) tell you about the flow of blood in this artery? (c) Make a table of values of \(v(r)\) for \(r=0,0.1,0.2,0.3\) \(0.4,0.5\) (image cannot copy)
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