/*! 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 46 \(y=(x+\ln x)^{3}\) where \(x=1\... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 46

\(y=(x+\ln x)^{3}\) where \(x=1\)

Short Answer

Expert verified
Given ;)

Step by step solution

01

Substitute the Given Value

Replace the variable x with the given value 1 in the function y. So the expression becomes ;)
02

Simplify the Expression Inside the Parentheses

First, evaluate the expression inside the parentheses:
03

Compute the Natural Logarithm

Calculate the natural logarithm of 1, . Since ;)
04

Simplify Further

Add the results from the previous step: . Hence, ,. In this case ;,)
05

Evaluate the Cubed Expression

Lastly, evaluate the cubed term: ;

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Key Concepts

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

Understanding the Natural Logarithm
Natural logarithms are logarithms with a base of e (approximately 2.718). They are written as \(\ln x\) and are essential in advanced mathematics.
Let's break it down:
  • When you calculate \(\ln x\), you're finding the power to which e must be raised to get x.
  • For example, \(\ln e = 1\) because e to the power of 1 is e.
Natural logarithms are widely used in calculus, particularly for finding derivatives and integrals of exponential functions.
Exploring Exponential Functions
Exponential functions involve numbers raised to a variable power, often represented as \(e^x\). These functions have unique properties:
  • The rate of change of \(e^x\) is proportional to its current value.
  • Its derivative is simply \(e^x\), making calculations straightforward.
Exponential functions appear frequently in growth and decay problems, as well as in financial mathematics.
Calculus Concepts in Detail
Calculus is the study of change and motion. It primarily involves derivatives and integrals, which help us understand how functions behave:
  • **Derivatives** measure how a function changes as its input changes.
  • **Integrals** compute the accumulation of quantities and can be thought of as the area under a curve.
In our exercise, we're using derivatives to find the rate of change of a function involving a natural logarithm and exponential terms. Here's the step-by-step reasoning:
  • We start with \(y=(x+\ln x)^3\).
  • Substituting x with 1, the expression becomes \(y=(1+\ln 1)^3\).
  • Since \(\ln 1=0\), we have \(y=(1+0)^3 = 1\).
This simplification demonstrates the function's behavior at that specific point.

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Most popular questions from this chapter

Use a graphing utility to draw the graphs of \(y=\sqrt{3^{x}}, y=\sqrt{3^{-x}}\), and \(y=3^{-x}\) on the same set of axes. How do these graphs differ? (Suggestion: Use the graphing window \([-4,4] 1\) by \([-2,6] 1\) )

\(y=\ln \sqrt{x^{2}+4 x+1}\)

Use a graphing utility to draw the graphs of \(y=3^{x}\) and \(y=4-\ln \sqrt{x}\) on the same axes. Then use TRACE and ZOOM to find all points of intersection of the two graphs.

In Exercises 39 through 42 , find the largest and small values of the given function over the prescribed closi bounded interval. 39\. \(f(x)=\ln \left(4 x-x^{2}\right) \quad\) for \(1 \leq x \leq 3\)

\(h(t)=\left(e^{-t}+e^{t}\right)^{5} \quad\) for \(-1 \leq t \leqq 1\)
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