Given #if f(x)=x^2-4# and #g(x)=sqrt(x-3)# how do you find f(g(x)) and g(f(x))?

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AA_123 Share
Dec 5, 2016

Answer:

#f(g(x))=x-7#

#g(f(x))=sqrt(x^2-7)#

Explanation:

Put #g# inside #f# to find #f(g(x))#

#f(x)=color(red)x^2-4#

#g(x)=color(red)sqrt(x-3)#

#f(g(x))=(color(red)sqrt(x-3))^2-4#

#color(white)(f(g(x)))=x-3-4#

#color(white)(f(g(x)))=x-7#

Put #f# inside #g# to find #g(f(x))#

#g(x)=sqrt(color(red)x-3)#

#f(x)=color(red)(x^2-4)#

#g(f(x))=sqrt(color(red)(x^2-4)-3)#

#color(white)(g(f(x)))=sqrt(x^2-7)#

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Write your answer here...
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Then teach the underlying concepts
Don't copy without citing sources
preview
?

Answer

Write a one sentence answer...

Answer:

Explanation

Explain in detail...

Explanation:

I want someone to double check my answer

Describe your changes (optional) 200

2
Dec 5, 2016

Answer:

The answers are #f(g(x))=x-7#
and #g(f(x))=sqrt(x^2-7)#

Explanation:

This is a composition of functions.

#f(x)=x^2-4#

#g(x)=sqrt(x-3)#

#fog(x)=f(g(x))=f(sqrt(x-3))=(sqrt(x-3))^2-4#

#=x-3-4=x-7#

And

#gof(x)=g(f(x))=g(x^2-4)=sqrt((x^2-4)-3)#

#=sqrt(x^2-4-3)=sqrt(x^2-7)#

You can see that

#f(g(x))!=#g(f(x))

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