/*! 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 18 \(h(x)=\sqrt[5]{2 x}\)... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 18

\(h(x)=\sqrt[5]{2 x}\)

Short Answer

Expert verified
The function \(h(x) = \sqrt[5]{2x}\) is defined and produces real number outputs for all real numbers including negative values, positive values, and zero. For example, \(h(-1) = -1.1487\), \(h(0) = 0\), and \(h(1) = 1.1487\).

Step by step solution

01

Identify the Function

The function given is \(h(x)=\sqrt[5]{2 x}\). This is a root function, specifically the fifth root of \(2x\). The root function is defined for all real numbers when the root is an odd number.
02

Evaluate the Function for a Positive Value

Let's evaluate the function at \(x=1\). So \(h(1) = \sqrt[5]{2 \cdot 1} = 1.1487\). This shows that the function is defined for positive values of \(x\) and takes on real number outputs.
03

Evaluate the Function for Zero

Evaluating the function where \(x=0\), we'll get \(h(0)=\sqrt[5]{2 \cdot 0} = 0\). This shows the function also works for zero input and will give an output of zero.
04

Evaluate the Function for a Negative Value

Finally, let's evaluate the function at \(x=-1\). We get \(h(-1) = \sqrt[5]{2 \cdot -1} = -1.1487\). This shows that the function is defined for negative numbers and gives real number outputs when the root is an odd number, in this case, five.

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 91影视!

Key Concepts

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

Real Number Outputs
When dealing with functions like the one given in the exercise, it's important to focus on what outputs we can expect. A function will map inputs to outputs. For the function \( h(x)=\sqrt[5]{2x} \), this involves a mathematical operation known as taking the fifth root.

One of the key points to understand is that the outputs of this function are real numbers. This means:
  • Any input from real numbers can lead to a real number output when considering odd roots, like a fifth root.
  • Even if \(x\) is negative, as long as the operation involves an odd root, the result is a real number.
  • This includes positive values, negative values, and zero.
Real number outputs ensure that the function is broad in terms of applicability to real-world contexts, which often deal with real numbers.
Root Function
A root function takes a number and performs a root operation on it. In this particular exercise, the focus is on the fifth root function, as expressed in \( h(x)=\sqrt[5]{2x} \).

The term 'fifth root' means that you are finding a number that, when multiplied by itself five times, will equal the original number. Some critical aspects of root functions include:
  • Odd root functions, like the fifth root, are defined for all real numbers.
  • These functions handle both negative and positive inputs differently compared to even roots.
  • Unlike square roots (even roots), which are not defined for negative numbers in the real number system, fifth root functions are friendly to negative inputs.
Understanding these properties helps in evaluating the behavior and the real-world implications of the function. Root functions are foundational in calculus and various mathematical analyses.
Function Evaluation
Evaluating a function involves calculating its output based on various given inputs. For the given function \( h(x)=\sqrt[5]{2x} \), evaluation helps in understanding the function's characteristics and expected outcomes.

When evaluating, the steps usually involve:
  • Substituting the value of \(x\) into the function.
  • Simplifying the expression, which may involve using a calculator for complex root calculations.
  • Interpreting the result, which illustrates a real and possibly insightful outcome regarding the behavior of the function.
For instance, evaluating at \(x=1\) results in approximately 1.1487. At \(x=0\), the result is a simple zero, and at \(x=-1\), the output is approximately -1.1487. Each of these evaluations shows how depending on the input, the output can vary widely, yet remains within the domain of real numbers. Function evaluation is crucial for applications in mathematics and various fields of engineering and science.

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

\(\frac{1}{4} y^2=x\)

\(h(x)=\sqrt{x^2-2 x}\)

Simplify the expression. Write your answer using only positive exponents. $$ (-3)^{-3} $$

\(f(x)=x^{1 / 3}, g(x)=\frac{1}{3} x^{1 / 3}+6\)

\(f(x)=\sqrt[3]{x}, g(x)=-\sqrt[3]{x}-1\)
See all solutions

Recommended explanations on Math Textbooks

Statistics

Read Explanation

Calculus

Read Explanation

Pure Maths

Read Explanation

Decision Maths

Read Explanation

Probability and Statistics

Read Explanation

Discrete Mathematics

Read Explanation
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.