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Mathematics 19 Online
OpenStudy (anonymous):

If f(x)=4x+7 and g(x)= sqrt(x-3), what is (fog)(4)

OpenStudy (anonymous):

keep getting these square root ones wrong. medal to correct answer

OpenStudy (lgbasallote):

lol bribing?

OpenStudy (anonymous):

well, just want people to double check there answer

OpenStudy (lgbasallote):

well what's your answer

OpenStudy (anonymous):

lemme do it

OpenStudy (lgbasallote):

tag me when you have an answer

OpenStudy (anonymous):

\[\sqrt{23-3}\] 4.4721 but thats not right.

OpenStudy (anonymous):

@lgbasallote

OpenStudy (lgbasallote):

solve for g(4) first

OpenStudy (lgbasallote):

\[(f o g)(4) \implies f(g(4))\]

OpenStudy (anonymous):

ahhh, so 11

OpenStudy (lgbasallote):

yup

OpenStudy (anonymous):

here is one that introduces h @lgbasallote Let f(x)=(4x^3+20)^2 and g(x)=4x^3+20. Given that f(x)=(hog)x, find h(x)

OpenStudy (anonymous):

where do i start with that

OpenStudy (lgbasallote):

find (h o g) x first

OpenStudy (vishweshshrimali5):

sorry to interrupt but Prefer this link .. for understanding ; http://www.regentsprep.org/Regents/math/algtrig/ATP7/compositionfunctions.htm

OpenStudy (vishweshshrimali5):

see the examples. . you may continue @lgbasallote

OpenStudy (anonymous):

this may help, in helping me

OpenStudy (anonymous):

wait, how do i find (hog)?

OpenStudy (lgbasallote):

find g(x) first

OpenStudy (lgbasallote):

what is g(x)

OpenStudy (anonymous):

what do i use for x?

OpenStudy (lgbasallote):

x

OpenStudy (anonymous):

isnt it just 4x^3+20 then?

OpenStudy (lgbasallote):

yup

OpenStudy (lgbasallote):

so therefore h(4x^3 + 20) = f(x) do you agree with this?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

where do i use that?

OpenStudy (lgbasallote):

hmm this was easier in my head...

OpenStudy (anonymous):

lol

OpenStudy (lgbasallote):

how will you turn h(4x^3 + 20) into h(x) what do you think?

OpenStudy (anonymous):

geeze, I have no clue.

OpenStudy (lgbasallote):

me neither :/

OpenStudy (lgbasallote):

try posting it as a new question maybe someone has an idea

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