Explanation Given: The given equation is e − x 2 = x Concept Used: The Intermediate Value Theorem: If f is continuous on the closed interval [ a , b ] and let N be any number between f ( a ) and f ( b ) , where f ( a ) ≠ f ( b ) . Then there exists a number c in ( a , b ) such that f ( c ) = N . Calculation: To show that there is a root of the equation e − x 2 = x in interval ( 0 , 1 ) First let f ( x ) = e − x 2 − x = 0 and a = 0 , b = 1

15th Edition

ISBN: 9781133109631

Author: Stewart, JAMES, Day, Troy

Publisher: Cengage Learning,

See similar textbooks

Concept explainers

Rate of Change

The relation between two quantities which displays how much greater one quantity is than another is called ratio.

Slope

The change in the vertical distances is known as the rise and the change in the horizontal distances is known as the run. So, the rise divided by run is nothing but a slope value. It is calculated with simple algebraic equations as:

Videos

Question

Chapter 2, Problem 34E

To determine

To show that there is a root of the equation:

e−x2=x , in interval (0,1) , using the Intermediate Value Theorem

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 2, Problem 33E

Chapter 2, Problem 1CS

Chapter 2 Solutions

Biocalculus

Ch. 2.1 - Prob. 1E Ch. 2.1 - Prob. 2E Ch. 2.1 - Prob. 3E Ch. 2.1 - Prob. 4E Ch. 2.1 - Prob. 5E Ch. 2.1 - Prob. 6E Ch. 2.1 - Prob. 7E Ch. 2.1 - Prob. 8E Ch. 2.1 - Prob. 9E Ch. 2.1 - Prob. 10E

Ch. 2.1 - Prob. 11E Ch. 2.1 - Prob. 12E Ch. 2.1 - Prob. 13E Ch. 2.1 - Prob. 14E Ch. 2.1 - Prob. 15E Ch. 2.1 - Prob. 16E Ch. 2.1 - Prob. 17E Ch. 2.1 - Prob. 18E Ch. 2.1 - Prob. 19E Ch. 2.1 - Prob. 20E Ch. 2.1 - Prob. 21E Ch. 2.1 - Prob. 22E Ch. 2.1 - Prob. 23E Ch. 2.1 - Prob. 24E Ch. 2.1 - Prob. 25E Ch. 2.1 - Prob. 26E Ch. 2.1 - Prob. 27E Ch. 2.1 - Prob. 28E Ch. 2.1 - Prob. 29E Ch. 2.1 - Prob. 30E Ch. 2.1 - Prob. 31E Ch. 2.1 - Prob. 32E Ch. 2.1 - Prob. 33E Ch. 2.1 - Prob. 34E Ch. 2.1 - Prob. 35E Ch. 2.1 - Prob. 36E Ch. 2.1 - Prob. 37E Ch. 2.1 - Prob. 38E Ch. 2.1 - Prob. 39E Ch. 2.1 - Prob. 40E Ch. 2.1 - Prob. 41E Ch. 2.1 - Prob. 42E Ch. 2.1 - Prob. 43E Ch. 2.1 - Prob. 44E Ch. 2.1 - Prob. 45E Ch. 2.1 - Prob. 46E Ch. 2.1 - Prob. 47E Ch. 2.1 - Prob. 48E Ch. 2.1 - Prob. 49E Ch. 2.1 - Prob. 50E Ch. 2.1 - Prob. 51E Ch. 2.1 - Prob. 52E Ch. 2.1 - Prob. 53E Ch. 2.1 - Prob. 54E Ch. 2.1 - Prob. 55E Ch. 2.1 - Prob. 56E Ch. 2.1 - Prob. 57E Ch. 2.1 - Prob. 58E Ch. 2.1 - Prob. 59E Ch. 2.1 - Prob. 60E Ch. 2.1 - Prob. 1P Ch. 2.1 - Prob. 2P Ch. 2.1 - Prob. 3P Ch. 2.1 - Prob. 4P Ch. 2.1 - Prob. 5P Ch. 2.1 - Prob. 6P Ch. 2.1 - Prob. 7P Ch. 2.1 - Prob. 8P Ch. 2.1 - Prob. 9P Ch. 2.2 - Prob. 1E Ch. 2.2 - Prob. 2E Ch. 2.2 - Prob. 3E Ch. 2.2 - Prob. 4E Ch. 2.2 - Prob. 5E Ch. 2.2 - Prob. 6E Ch. 2.2 - Prob. 7E Ch. 2.2 - Prob. 8E Ch. 2.2 - Prob. 9E Ch. 2.2 - Prob. 10E Ch. 2.2 - Prob. 11E Ch. 2.2 - Prob. 12E Ch. 2.2 - Prob. 13E Ch. 2.2 - Prob. 14E Ch. 2.2 - Prob. 15E Ch. 2.2 - Prob. 16E Ch. 2.2 - Prob. 17E Ch. 2.2 - Prob. 18E Ch. 2.2 - Prob. 19E Ch. 2.2 - Prob. 20E Ch. 2.2 - Prob. 21E Ch. 2.2 - Prob. 22E Ch. 2.2 - Prob. 23E Ch. 2.2 - Prob. 24E Ch. 2.2 - Prob. 25E Ch. 2.2 - Prob. 26E Ch. 2.2 - Prob. 27E Ch. 2.2 - Prob. 28E Ch. 2.2 - Prob. 29E Ch. 2.2 - Prob. 30E Ch. 2.2 - Prob. 31E Ch. 2.2 - Prob. 32E Ch. 2.2 - Prob. 33E Ch. 2.2 - Prob. 34E Ch. 2.2 - Prob. 35E Ch. 2.2 - Prob. 36E Ch. 2.2 - Prob. 37E Ch. 2.3 - Prob. 1E Ch. 2.3 - Prob. 2E Ch. 2.3 - Prob. 3E Ch. 2.3 - Prob. 4E Ch. 2.3 - Prob. 5E Ch. 2.3 - Prob. 6E Ch. 2.3 - Prob. 7E Ch. 2.3 - Prob. 8E Ch. 2.3 - Prob. 9E Ch. 2.3 - Prob. 10E Ch. 2.3 - Prob. 11E Ch. 2.3 - Prob. 12E Ch. 2.3 - Prob. 13E Ch. 2.3 - Prob. 14E Ch. 2.3 - Prob. 15E Ch. 2.3 - Prob. 16E Ch. 2.3 - Prob. 17E Ch. 2.3 - Prob. 18E Ch. 2.3 - Prob. 19E Ch. 2.3 - Prob. 20E Ch. 2.3 - Prob. 21E Ch. 2.3 - Prob. 22E Ch. 2.3 - Prob. 23E Ch. 2.3 - Prob. 24E Ch. 2.3 - Prob. 25E Ch. 2.3 - Prob. 26E Ch. 2.3 - Prob. 27E Ch. 2.3 - Prob. 28E Ch. 2.3 - Prob. 29E Ch. 2.3 - Prob. 30E Ch. 2.3 - Prob. 31E Ch. 2.3 - Prob. 32E Ch. 2.3 - Prob. 33E Ch. 2.3 - Prob. 34E Ch. 2.3 - Prob. 35E Ch. 2.3 - Prob. 36E Ch. 2.3 - Prob. 37E Ch. 2.3 - Prob. 38E Ch. 2.3 - Prob. 39E Ch. 2.3 - Prob. 40E Ch. 2.3 - Prob. 41E Ch. 2.3 - Prob. 42E Ch. 2.3 - Prob. 43E Ch. 2.3 - Prob. 44E Ch. 2.4 - Prob. 1E Ch. 2.4 - Prob. 2E Ch. 2.4 - Prob. 3E Ch. 2.4 - Prob. 4E Ch. 2.4 - Prob. 5E Ch. 2.4 - Prob. 6E Ch. 2.4 - Prob. 7E Ch. 2.4 - Prob. 8E Ch. 2.4 - Prob. 9E Ch. 2.4 - Prob. 10E Ch. 2.4 - Prob. 11E Ch. 2.4 - Prob. 12E Ch. 2.4 - Prob. 13E Ch. 2.4 - Prob. 14E Ch. 2.4 - Prob. 15E Ch. 2.4 - Prob. 16E Ch. 2.4 - Prob. 17E Ch. 2.4 - Prob. 18E Ch. 2.4 - Prob. 19E Ch. 2.4 - Prob. 20E Ch. 2.4 - Prob. 21E Ch. 2.4 - Prob. 22E Ch. 2.4 - Prob. 23E Ch. 2.4 - Prob. 24E Ch. 2.4 - Prob. 25E Ch. 2.4 - Prob. 26E Ch. 2.4 - Prob. 27E Ch. 2.4 - Prob. 28E Ch. 2.4 - Prob. 29E Ch. 2.4 - Prob. 30E Ch. 2.4 - Prob. 31E Ch. 2.4 - Prob. 32E Ch. 2.4 - Prob. 33E Ch. 2.4 - Prob. 34E Ch. 2.4 - Prob. 35E Ch. 2.4 - Prob. 36E Ch. 2.4 - Prob. 37E Ch. 2.4 - Prob. 38E Ch. 2.4 - Prob. 39E Ch. 2.4 - Prob. 40E Ch. 2.4 - Prob. 41E Ch. 2.4 - Prob. 42E Ch. 2.4 - Prob. 43E Ch. 2.4 - Prob. 44E Ch. 2.4 - Prob. 45E Ch. 2.4 - Prob. 46E Ch. 2.4 - Prob. 47E Ch. 2.4 - Prob. 48E Ch. 2.4 - Prob. 49E Ch. 2.4 - Prob. 50E Ch. 2.4 - Prob. 51E Ch. 2.4 - Prob. 52E Ch. 2.5 - Prob. 1E Ch. 2.5 - Prob. 2E Ch. 2.5 - Prob. 3E Ch. 2.5 - Prob. 4E Ch. 2.5 - Prob. 5E Ch. 2.5 - Prob. 6E Ch. 2.5 - Prob. 7E Ch. 2.5 - Prob. 8E Ch. 2.5 - Prob. 9E Ch. 2.5 - Prob. 10E Ch. 2.5 - Prob. 11E Ch. 2.5 - Prob. 12E Ch. 2.5 - Prob. 13E Ch. 2.5 - Prob. 14E Ch. 2.5 - Prob. 15E Ch. 2.5 - Prob. 16E Ch. 2.5 - Prob. 17E Ch. 2.5 - Prob. 18E Ch. 2.5 - Prob. 19E Ch. 2.5 - Prob. 20E Ch. 2.5 - Prob. 21E Ch. 2.5 - Prob. 22E Ch. 2.5 - Prob. 23E Ch. 2.5 - Prob. 24E Ch. 2.5 - Prob. 25E Ch. 2.5 - Prob. 26E Ch. 2.5 - Prob. 27E Ch. 2.5 - Prob. 28E Ch. 2.5 - Prob. 29E Ch. 2.5 - Prob. 30E Ch. 2.5 - Prob. 31E Ch. 2.5 - Prob. 32E Ch. 2.5 - Prob. 33E Ch. 2.5 - Prob. 34E Ch. 2.5 - Prob. 35E Ch. 2.5 - Prob. 36E Ch. 2.5 - Prob. 37E Ch. 2.5 - Prob. 38E Ch. 2.5 - Prob. 39E Ch. 2.5 - Prob. 40E Ch. 2.5 - Prob. 41E Ch. 2.5 - Prob. 42E Ch. 2.5 - Prob. 43E Ch. 2.5 - Prob. 44E Ch. 2.5 - Prob. 45E Ch. 2.5 - Prob. 46E Ch. 2.5 - Prob. 47E Ch. 2.5 - Prob. 48E Ch. 2.5 - Prob. 49E Ch. 2.5 - Prob. 50E Ch. 2.5 - Prob. 51E Ch. 2 - Prob. 1CC Ch. 2 - Prob. 2CC Ch. 2 - Prob. 3CC Ch. 2 - Prob. 4CC Ch. 2 - Prob. 5CC Ch. 2 - Prob. 6CC Ch. 2 - Prob. 7CC Ch. 2 - Prob. 8CC Ch. 2 - Prob. 9CC Ch. 2 - Prob. 10CC Ch. 2 - Prob. 1TFQ Ch. 2 - Prob. 2TFQ Ch. 2 - Prob. 3TFQ Ch. 2 - Prob. 4TFQ Ch. 2 - Prob. 5TFQ Ch. 2 - Prob. 6TFQ Ch. 2 - Prob. 7TFQ Ch. 2 - Prob. 8TFQ Ch. 2 - Prob. 9TFQ Ch. 2 - Prob. 10TFQ Ch. 2 - Prob. 11TFQ Ch. 2 - Prob. 12TFQ Ch. 2 - Prob. 13TFQ Ch. 2 - Prob. 14TFQ Ch. 2 - Prob. 15TFQ Ch. 2 - Prob. 16TFQ Ch. 2 - Prob. 1E Ch. 2 - Prob. 2E Ch. 2 - Prob. 3E Ch. 2 - Prob. 4E Ch. 2 - Prob. 5E Ch. 2 - Prob. 6E Ch. 2 - Prob. 7E Ch. 2 - Prob. 8E Ch. 2 - Prob. 9E Ch. 2 - Prob. 10E Ch. 2 - Prob. 11E Ch. 2 - Prob. 12E Ch. 2 - Prob. 13E Ch. 2 - Prob. 14E Ch. 2 - Prob. 15E Ch. 2 - Prob. 16E Ch. 2 - Prob. 17E Ch. 2 - Prob. 18E Ch. 2 - Prob. 19E Ch. 2 - Prob. 20E Ch. 2 - Prob. 21E Ch. 2 - Prob. 22E Ch. 2 - Prob. 23E Ch. 2 - Prob. 24E Ch. 2 - Prob. 25E Ch. 2 - Prob. 26E Ch. 2 - Prob. 27E Ch. 2 - Prob. 28E Ch. 2 - Prob. 29E Ch. 2 - Prob. 30E Ch. 2 - Prob. 31E Ch. 2 - Prob. 32E Ch. 2 - Prob. 33E Ch. 2 - Prob. 34E Ch. 2 - Prob. 1CS Ch. 2 - Prob. 2CS Ch. 2 - Prob. 3CS Ch. 2 - Prob. 4CS Ch. 2 - Prob. 5CS Ch. 2 - Prob. 6CS Ch. 2 - Prob. 7CS Ch. 2 - Prob. 8CS Ch. 2 - Prob. 9CS

Knowledge Booster

$Calculus$

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, calculus and related others by exploring similar questions and additional content below.

To show that there is a root of the equation: e − x 2 = x , in interval ( 0 , 1 ) , using the Intermediate Value Theorem

Solution Summary: The author calculates the root of the equation e-x2=x using the Intermediate Value Theorem.

Concept explainers

Videos

To show that there is a root of the equation:

e−x2=x , in interval (0,1) , using the Intermediate Value Theorem

Want to see the full answer?

Chapter 2 Solutions

To show that there is a root of the equation: e − x 2 = x , in interval ( 0 , 1 ) , using the Intermediate Value Theorem

Solution Summary: The author calculates the root of the equation e-x2=x using the Intermediate Value Theorem.

Concept explainers

Videos

To show that there is a root of the equation: e−x2=x , in interval (0,1) , using the Intermediate Value Theorem

Want to see the full answer?

Chapter 2 Solutions

To show that there is a root of the equation:

e−x2=x , in interval (0,1) , using the Intermediate Value Theorem