a. Explanation Concept Used: In this we have to use the concept of function,domain and range. Calculation: D f : ( 0 , + ∞ ) Domain of f: ln ( x ) is defined for x > 0 D g : ( − ∞ , + ∞ ) Domain of g: x 2 - 9 is defined for any x in the set of Real numbers R. ( f ∘ g ) ( x ) = f ( g ( x ) ) = f ( x 2 − 9 ) = ln ( x 2 − 9 ) x 2 − 9 > 0 → x 2 > 9 → x 2 > 9 → | x | > 3 → ( − ∞ , − 3 b. To determine To find the function and domain of g ∘ f ,if f ( x ) = ln ( x ) and g ( x ) = x 2 − 9 . c. To determine To find the function and domain of f ∘ f ,if f ( x ) = ln ( x ) and g ( x ) = x 2 − 9 . d. To determine Find the function and domain of g ∘ g ,if f ( x ) = ln ( x ) and g ( x ) = x 2 − 9 .

15th Edition

ISBN: 9781305177291

Author: James Stewart, Troy Day

Publisher: Cengage Learning

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Differential Equations

Have you ever observed the movement of the satellites and rockets or how the planets go around the sun in their orbits? How will you determine their motion? Definitely they follow some scientific laws and these kinds of problems are framed as differentia…

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Chapter 1, Problem 21E

To determine

To find the function and domain of f∘g ,if f(x)=ln(x) and g(x)=x2−9 .

To determine

To find the function and domain of g∘f ,if f(x)=ln(x) and g(x)=x2−9 .

To determine

To find the function and domain of f∘f ,if f(x)=ln(x) and g(x)=x2−9 .

To determine

Find the function and domain of g∘g ,if f(x)=ln(x) and g(x)=x2−9 .

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Chapter 1, Problem 20E

Chapter 1, Problem 22E

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Chapter 1 Solutions

Biocalculus: Calculus For Life Sciences

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To find the function and domain of f ∘ g ,if f ( x ) = ln ( x ) and g ( x ) = x 2 − 9 .

Solution Summary: The author explains the concept of function,domain, and range in fcirc g.

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To find the function and domain of f∘g ,if f(x)=ln(x) and g(x)=x2−9 .

To find the function and domain of g∘f ,if f(x)=ln(x) and g(x)=x2−9 .

To find the function and domain of f∘f ,if f(x)=ln(x) and g(x)=x2−9 .

Find the function and domain of g∘g ,if f(x)=ln(x) and g(x)=x2−9 .

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Chapter 1 Solutions