Question

Given that f(x) = x2 + 3x + 6 and g(x) = , solve for f(g(x)) when x = 1.

a. -1
b. 0
c. 3
d. 4

please walk through/explain it, I want to understand it!!

Answer 1

missing something?

Answer 2

what do you mean?

Answer 3

@satellite73

Answer 4

look carefully at your question
i think there is a copy and paste fail

Answer 5

oh oops its G(x) = (x-3 / 2)

Answer 6

\[g(x)=\frac{x-3}{2}\]?

Answer 7

yes

Answer 8

k
step 1
find
\[g(1)\]

Answer 9

you good with that?

Answer 10

Uh no I think I've learned it a different way than you, I thought you plugged in g(x) into the f(x) equation

Answer 11

you are being asked for
\[f(g(1))\]right?

Answer 12

yes sorry my bad

Answer 13

k then you need \(g(1)\) first

Answer 14

so would it be F(x) = (\frac{ 1-3 }{ 2 })^{2} +3(\frac{ 1-3 }{ 2 } )+ 6\]

Answer 15

\[ F(x) = (\frac{ 1-3 }{ 2 })^{2} +3(\frac{ 1-3 }{ 2 } )+ 6\] is way way way too much work and although it is correct, not to be rued but it is rather silly
lets take it step by step

Answer 16

\[g(x)=\frac{x-3}{2}, g(1)=\frac{1-3}{2}=-1\]
\[f(x)=x^2+3x+6, f(-1)=(-1)^2+3\times (-1)+6=4\]

Answer 17

So then the answer would be 3???

Answer 18

i get 4

Answer 19

\[(-1)^2+3\times (-1)+6=1-3+6=-2+6=4\]

Answer 20

Oh nevermind I get it now!! Thank you!!!

Answer 21

yw

Answer 22

now if you had wanted
\[f(g(x))\] you would have had to do what you did and write
\[f(g(x))=(\frac{ x-3 }{ 2 })^{2} +3(\frac{ x-3 }{ 2 } )+ 6\]

Answer 23

so technically I was correct

Answer 24

of course

Answer 25

@satellite73 okay thank you! do you think you could help me with this other one I'm stuck on?