Question

Given that f(x) = x2 + 3x + 6 and g(x) = x-3 over 2, solve for f(g(x)) when x = 1 SOMEBODY PLEASE EXPLAIN HOW TO SOLVE

Answer 1

what is g(1) ?

Answer 2

Given that f(x) = x2 + 3x + 6 and g(x) = the quantity of x plus two, over three, solve for f(g(x)) when x = 1

Answer 3

thats the question

Answer 4

no sorry i understand the question, i am asking you if you can evaluate g(1)
the goal is to find f(g(1))

Answer 5

thats where im confused
idk where to start..

Answer 6

if i say h(x) = x - 5 , evaluate h(3)

can you do that?

Answer 7

h(3) = 3 - 5 ?

Answer 8

yep good , which is -2
h(3) = -2

do same thing with function g(x) at x=1
g(1) = ?

Answer 9

g(1) = 1-5 ?
what is (g)

Answer 10

g is name of the function

g(x) is defined in the problem, i used the letter "h" because i made up a new function

Answer 11

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

Answer 12

** sorry (x-3)/2 whatever you know what it is

Answer 13

ok

Answer 14

g(1) = functoin 1 ?

Answer 15

??

Answer 16

im so confused

Answer 17

why?
just plug 1 in for x into g(x)

ok so you get some number g(1)
then you plug than number into f(x) to get f(g(1))
this new number is the answer

Answer 18

so g(1) = 1 ?

Answer 19

not sure you confused me about what g(x) is lol

Answer 20

g(x) = 1 right

Answer 21

yes if
g(x) = (x+2)/3

Answer 22

Given that f(x) = x2 + 3x + 6 and g(x) = the quantity of x plus two, over three, solve for f(g(x)) when x = 1

Answer 23

so how do i do this one

Answer 24

haha you are doing it
ok
g(1) = 1

now evaluate f(1)

Answer 25

1

Answer 26

f(1) = 1

Answer 27

nope

f(x) = x^2 +3x +6

f(1) = 1^2 +3(1) +6