Ask your own question, for FREE!
Mathematics 7 Online
OpenStudy (anonymous):

Find (h.f.g)(-2) if f(x) = x2 - 5, g(x) = 3x + 1, h(x) = x + 3

OpenStudy (anonymous):

is that hxfxg or composite functions?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

yes to which one?

OpenStudy (anonymous):

composite

OpenStudy (anonymous):

ok, well, when you have a composite function, you work from the inside out. First figure out the innermost value (which would be f(-2). Then plug that into g(x) and then plug the value you got for g(x) into f(x)

OpenStudy (anonymous):

im really lost with this could u get me the answer plz that would be greatly appricated

OpenStudy (anonymous):

well, first calculate h(-2)=(-2)+3 Then calculate g(h(-2) And then f(g(h(-2)) :D

OpenStudy (anonymous):

im still not gettin it

OpenStudy (anonymous):

i might have told u the wrong thing this is what i am looking for (hfg)(-2) =

OpenStudy (anonymous):

:? what about it? ok-- f(g(h(-2))))=f(g(-2+3))=f(g(1))=f(3(1)+1))=f(4)=4^2 -5=9

OpenStudy (anonymous):

Oh, sorry, my bad ok, we just switch it around a little

OpenStudy (anonymous):

h(f(g((-2))))=h(f(3(-2)+1)))=h(f(-5))=h(25-5)=h(20)=20+3=23

OpenStudy (anonymous):

THANK U

OpenStudy (anonymous):

:D but do u understand how to do solve it urself?

OpenStudy (anonymous):

YES I THINK SO ITS STILL KINDA HARD THO

OpenStudy (anonymous):

if u get another question like it i can try helping u understand :)

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
Arriyanalol: i need help
16 hours ago 7 Replies 1 Medal
Arriyanalol: CONGRATS to me
19 hours ago 8 Replies 2 Medals
Legomaster346: What colleges are good that are near Arab Alabama?
1 week ago 5 Replies 0 Medals
Arriyanalol: i need help
1 week ago 6 Replies 2 Medals
Arizona: does anyone know the percentage rate of 30,000 and 40,000
1 week ago 0 Replies 0 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!