Calculus 2012 Student Edition (by Finney/Demana/Waits/Kennedy)

4th Edition

ISBN: 9780133178579

Author: Ross L. Finney

Publisher: PEARSON

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Equations and inequalities describe the relationship between two mathematical expressions.

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A linear function can just be a constant, or it can be the constant multiplied with the variable like x or y. If the variables are of the form, x2, x1/2 or y2 it is not linear. The exponent over the variables should always be 1.

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Question

Chapter 5.2, Problem 49E

To determine

To show : That the equation has exactly one solution in the interval.

Expert Solution & Answer

Answer to Problem 49E

The possible explanation is given below.

Explanation of Solution

Given information :

The given equation is $x+ln(x+1)=0, 0≤x≤3$

Calculation :

Let $f(x)=x+ln(x+1)$ .

Substitute $x=0$ and solve for $f(x)$ :

$f(0)=0+ln(0+1)f(0)=ln(1)f(0)=0$

Since, it is known that $f(x)$ has at least one solution on the interval $[0, 3]$ .

To prove that $f(x)$ has exactly one solution use proof by contradiction.

Since, $f(x)$ is continuous on the interval $[0, 3]$ and differentiable on the interval $(0, 3)$ .

Suppose, that $f(x)$ has one more solution for $f(x)=0$ on the interval $[0, 3]$ , where $x=0$ is one solution and $x=t,$ where $t=0$ is another.

Therefore, $f(0)=f(a)=0$ .

The slope between the two point is

$f(a)−f(0)a−0=0a−0f(a)−f(0)a−0=0$

Thus, by Mean Value Theorem, there must be a $c$ in the interval $(0, a)$ such that $f'(c)=0$ .

However, $f'(x)=1+1x+1>0$ in the interval in the interval $[0, 3]$ , which means $f'(c)≠0$ for all $c$ in $(0, 3)$ . This contradicts the fact that $f'(c)=0$ for $(0, a)$ .

Therefore ,

$f(x)$ cannot have a second solution for $f(x)=0$ so $x+ln(x+1)=0$ has exactly one solution.

Chapter 5.2, Problem 48E

Chapter 5.2, Problem 50E

Chapter 5 Solutions

Calculus 2012 Student Edition (by Finney/Demana/Waits/Kennedy)

Ch. 5.1 - Prob. 1QR Ch. 5.1 - Prob. 2QR Ch. 5.1 - Prob. 3QR Ch. 5.1 - Prob. 4QR Ch. 5.1 - Prob. 5QR Ch. 5.1 - Prob. 6QR Ch. 5.1 - Prob. 7QR Ch. 5.1 - Prob. 8QR Ch. 5.1 - Prob. 9QR Ch. 5.1 - Prob. 10QR

Ch. 5.1 - Prob. 11QR Ch. 5.1 - Prob. 12QR Ch. 5.1 - Prob. 1E Ch. 5.1 - Prob. 2E Ch. 5.1 - Prob. 3E Ch. 5.1 - Prob. 4E Ch. 5.1 - Prob. 5E Ch. 5.1 - Prob. 6E Ch. 5.1 - Prob. 7E Ch. 5.1 - Prob. 8E Ch. 5.1 - Prob. 9E Ch. 5.1 - Prob. 10E Ch. 5.1 - Prob. 11E Ch. 5.1 - Prob. 12E Ch. 5.1 - Prob. 13E Ch. 5.1 - Prob. 14E Ch. 5.1 - Prob. 15E Ch. 5.1 - Prob. 16E Ch. 5.1 - Prob. 17E Ch. 5.1 - Prob. 18E Ch. 5.1 - Prob. 19E Ch. 5.1 - Prob. 20E Ch. 5.1 - Prob. 21E Ch. 5.1 - Prob. 22E Ch. 5.1 - Prob. 23E Ch. 5.1 - Prob. 24E Ch. 5.1 - Prob. 25E Ch. 5.1 - Prob. 26E Ch. 5.1 - Prob. 27E Ch. 5.1 - Prob. 28E Ch. 5.1 - Prob. 29E Ch. 5.1 - Prob. 30E Ch. 5.1 - Prob. 31E Ch. 5.1 - Prob. 32E Ch. 5.1 - Prob. 33E Ch. 5.1 - Prob. 34E Ch. 5.1 - Prob. 35E Ch. 5.1 - Prob. 36E Ch. 5.1 - Prob. 37E Ch. 5.1 - Prob. 38E Ch. 5.1 - Prob. 39E Ch. 5.1 - Prob. 40E Ch. 5.1 - Prob. 41E Ch. 5.1 - Prob. 42E Ch. 5.1 - Prob. 43E Ch. 5.1 - Prob. 44E Ch. 5.1 - Prob. 45E Ch. 5.1 - Prob. 46E Ch. 5.1 - Prob. 47E Ch. 5.1 - Prob. 48E Ch. 5.1 - 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