Ask your own question, for FREE!
Mathematics 19 Online
OpenStudy (anonymous):

If f(x)=x+7 and g(x)=x-7 (a) f(g(x)=_______ (b) g(f(x)=_______ (c) thus g(x) is called an inverse function of f(x) CAN I GET HELP ON THIS?

OpenStudy (anonymous):

f(g(x)) = g(x) + 7 = (x-7) + 7 = ?

OpenStudy (anonymous):

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

OpenStudy (anonymous):

so f(g(x))= (x-7)+7

OpenStudy (anonymous):

for f(g(x)), you plug in g(x) into f(x)

OpenStudy (anonymous):

You can then remove the ( ) and simplify further.

OpenStudy (anonymous):

And part c isnt really a problem, its just true

OpenStudy (anonymous):

i got (c) right, it is called an inverse, (a) is then (x-7)(x+7) and converts to x^2-49 then?

OpenStudy (anonymous):

thats f(x)*g(x)

OpenStudy (anonymous):

but yea your (x-7)(x+7) = x^2 -49

OpenStudy (anonymous):

cause i typed in that last part into my web homework and got it wrong, it isnt x^2-49

OpenStudy (anonymous):

thats because part a asks for f(g(x)). thats different than f(x)*g(x)

OpenStudy (anonymous):

so (a) is then (x-7)+7 and (b) is then (x+7)-7 ??

OpenStudy (anonymous):

yes. and they both are equal to x

OpenStudy (anonymous):

ok thank you a bunch

OpenStudy (anonymous):

no problem

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Latest Questions
Breathless: Spooky witch but cute
3 hours ago 3 Replies 0 Medals
Arriyanalol: help
3 hours ago 10 Replies 2 Medals
Arriyanalol: @tinydinoUwU stop trying to find a argument u blad lil boy
1 day ago 5 Replies 3 Medals
Jaded012023: Please tell me what you all think of this song
6 hours ago 6 Replies 1 Medal
Arriyanalol: bro how
6 hours ago 2 Replies 3 Medals
Arriyanalol: cant wait for the new bluey movie in 2027
1 day ago 12 Replies 2 Medals
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!