(a) Answer The real solutionsof the equation f ( x ) = x 2 − 5 for f ( x ) = 4 are x = 3 and x = − 3 . Explanation Given information: The equation is f ( x ) = x 2 − 5 and f ( x ) = 4 . Calculation: Substitute 4 for f ( x ) in the given equation and solve. x 2 − 5 = 4 x 2 = 9 x = ± 9 x = ± 3 Therefore, the real solution of the equation f ( x ) = x 2 − 5 for f ( x ) = 4 are x = 3 and x = − 3 . (b) To determine To find: The real solution of the equation f ( x ) = x 2 − 5 for f ( x ) = − 6 . Answer There is no real solution of the equation f ( x ) = x 2 − 5 for f ( x ) = − 6 . Explanation Given information: The equation is f ( x ) = x 2 − 5 and f ( x ) = − 6 . Calculation: Substitute − 6 for f ( x ) in the given equation and solve. x 2 − 5 = − 6 x 2 − 5 + 5 = − 6 + 5 x 2 = − 1 This is not possible for any real value of x . Therefore, there is no real solution of the equation f ( x ) = x 2 − 5 for f ( x ) = − 6 .

Calculus: Graphical, Numerical, Algebraic: Solutions Manual

3rd Edition

ISBN: 9780132014144

Author: Ross L. Finney, Franklin Demana, Bert K. Waits, Daniel Kennedy

Publisher: Pearson Prentice Hall

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Question

Chapter 1.2, Problem 9QR

(a)

To determine

To find:The real solution of the equation f(x)=x2−5 for f(x)=4 .

(a)

Expert Solution

Answer to Problem 9QR

The real solutionsof the equation f(x)=x2−5 for f(x)=4 are x=3 and x=−3 .

Explanation of Solution

Given information:The equation is f(x)=x2−5 and f(x)=4 .

Calculation:

Substitute 4 for f(x) in the given equation and solve.

x2−5=4x2=9x=±9x=±3

Therefore, the real solution of the equation f(x)=x2−5 for f(x)=4 are x=3 and x=−3 .

(b)

To determine

To find:The real solution of the equation f(x)=x2−5 for f(x)=−6 .

(b)

Expert Solution

Answer to Problem 9QR

There is no real solution of the equation f(x)=x2−5 for f(x)=−6 .

Explanation of Solution

Given information:The equation is f(x)=x2−5 and f(x)=−6 .

Calculation:

Substitute −6 for f(x) in the given equation and solve.

x2−5=−6x2−5+5=−6+5x2=−1

This is not possible for any real value of x .

Therefore, there is no real solution of the equation f(x)=x2−5 for f(x)=−6 .

Chapter 1.2, Problem 8QR

Chapter 1.2, Problem 10QR

Chapter 1 Solutions

Calculus: Graphical, Numerical, Algebraic: Solutions Manual

Ch. 1.1 - Prob. 1QR Ch. 1.1 - Prob. 2QR Ch. 1.1 - Prob. 3QR Ch. 1.1 - Prob. 4QR Ch. 1.1 - Prob. 5QR Ch. 1.1 - Prob. 6QR Ch. 1.1 - Prob. 7QR Ch. 1.1 - Prob. 8QR Ch. 1.1 - Prob. 9QR Ch. 1.1 - Prob. 10QR

Ch. 1.1 - Prob. 1E Ch. 1.1 - Prob. 2E Ch. 1.1 - Prob. 3E Ch. 1.1 - Prob. 4E Ch. 1.1 - Prob. 5E Ch. 1.1 - Prob. 6E Ch. 1.1 - Prob. 7E Ch. 1.1 - Prob. 8E Ch. 1.1 - Prob. 9E Ch. 1.1 - Prob. 10E Ch. 1.1 - Prob. 11E Ch. 1.1 - Prob. 12E Ch. 1.1 - Prob. 13E Ch. 1.1 - Prob. 14E Ch. 1.1 - Prob. 15E Ch. 1.1 - Prob. 16E Ch. 1.1 - Prob. 17E Ch. 1.1 - Prob. 18E Ch. 1.1 - Prob. 19E Ch. 1.1 - Prob. 20E Ch. 1.1 - Prob. 21E Ch. 1.1 - Prob. 22E Ch. 1.1 - Prob. 23E Ch. 1.1 - Prob. 24E Ch. 1.1 - Prob. 25E Ch. 1.1 - Prob. 26E Ch. 1.1 - Prob. 27E Ch. 1.1 - Prob. 28E Ch. 1.1 - Prob. 29E Ch. 1.1 - Prob. 30E Ch. 1.1 - Prob. 31E Ch. 1.1 - Prob. 32E Ch. 1.1 - Prob. 33E Ch. 1.1 - Prob. 34E Ch. 1.1 - Prob. 35E Ch. 1.1 - Prob. 36E Ch. 1.1 - Prob. 37E Ch. 1.1 - Prob. 38E Ch. 1.1 - Prob. 39E Ch. 1.1 - Prob. 40E Ch. 1.1 - Prob. 41E Ch. 1.1 - Prob. 42E Ch. 1.1 - Prob. 43E Ch. 1.1 - Prob. 44E Ch. 1.1 - Prob. 45E Ch. 1.1 - Prob. 46E Ch. 1.1 - Prob. 47E Ch. 1.1 - Prob. 48E Ch. 1.1 - Prob. 49E Ch. 1.1 - Prob. 50E Ch. 1.1 - 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Prob. 61E Ch. 1.6 - Prob. 62E Ch. 1.6 - Prob. 1QQ Ch. 1.6 - Prob. 2QQ Ch. 1.6 - Prob. 3QQ Ch. 1.6 - Prob. 4QQ Ch. 1 - Prob. 1RE Ch. 1 - Prob. 2RE Ch. 1 - Prob. 3RE Ch. 1 - Prob. 4RE Ch. 1 - Prob. 5RE Ch. 1 - Prob. 6RE Ch. 1 - Prob. 7RE Ch. 1 - Prob. 8RE Ch. 1 - Prob. 9RE Ch. 1 - Prob. 10RE Ch. 1 - Prob. 11RE Ch. 1 - Prob. 12RE Ch. 1 - Prob. 13RE Ch. 1 - Prob. 14RE Ch. 1 - Prob. 15RE Ch. 1 - Prob. 16RE Ch. 1 - Prob. 17RE Ch. 1 - Prob. 18RE Ch. 1 - Prob. 19RE Ch. 1 - Prob. 20RE Ch. 1 - Prob. 21RE Ch. 1 - Prob. 22RE Ch. 1 - Prob. 23RE Ch. 1 - Prob. 24RE Ch. 1 - Prob. 25RE Ch. 1 - Prob. 26RE Ch. 1 - Prob. 27RE Ch. 1 - Prob. 28RE Ch. 1 - Prob. 29RE Ch. 1 - Prob. 30RE Ch. 1 - Prob. 31RE Ch. 1 - Prob. 32RE Ch. 1 - Prob. 33RE Ch. 1 - Prob. 34RE Ch. 1 - Prob. 35RE Ch. 1 - Prob. 36RE Ch. 1 - Prob. 37RE Ch. 1 - Prob. 38RE Ch. 1 - Prob. 39RE Ch. 1 - Prob. 40RE Ch. 1 - Prob. 41RE Ch. 1 - Prob. 42RE Ch. 1 - Prob. 43RE Ch. 1 - Prob. 44RE Ch. 1 - Prob. 45RE Ch. 1 - Prob. 46RE Ch. 1 - Prob. 47RE Ch. 1 - 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The real solution of the equation f ( x ) = x 2 − 5 for f ( x ) = 4 .

Videos

To find:The real solution of the equation f(x)=x2−5 for f(x)=4 .

Answer to Problem 9QR

Explanation of Solution

To find:The real solution of the equation f(x)=x2−5 for f(x)=−6 .

Answer to Problem 9QR

Explanation of Solution

Chapter 1 Solutions

Additional Math Textbook Solutions