geez, what's with all the words?
f(x) = x^2 + 3x + 6
g(x) = (x-3)/2
g(1) = (1-3)/2 = -1
f(-1) = (-1)^2+3(-1)+6 = 1-3+6 = 4
or,
f(g(x)) = g^2+3g+6
= ((x-3)/2)^2 + 3(x-3)/2 + 6
= (x^2+15)/4
f(g(-1)) = ((-1)^2+15)/4 = 16/4 = 4
Given that f(x) = x^2 + 3x + 6 and g(x) = the quantity of x minus three, over two, solve for f(g(x)) when x = 1.
A -1
B 0
C 3
D 4
I say it is 4. Not sure though.
2 answers
easy way
g = [(x-3)/2 ] ???
g(1) = [(1-3)/2 ] = -1
f(-1) = 1-3+6 = 4
---------
hard way
f(g)=[(x-3)/2]^2+3[(x-3)/2 ]+6
put in x = 1
f(g)=[(1-3)/2]^2+3[(1-3)/2 ]+6
= 1 -3 +6
= 4
g = [(x-3)/2 ] ???
g(1) = [(1-3)/2 ] = -1
f(-1) = 1-3+6 = 4
---------
hard way
f(g)=[(x-3)/2]^2+3[(x-3)/2 ]+6
put in x = 1
f(g)=[(1-3)/2]^2+3[(1-3)/2 ]+6
= 1 -3 +6
= 4