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.

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

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

To solve for f(g(x)), we first need to find the value of g(x) when x = 1.

Given that g(x) = (x - 3) / 2, we substitute x = 1 into the expression for g(x):

g(1) = (1 - 3) / 2

Simplifying further:

g(1) = -2 / 2
g(1) = -1

Now that we have the value of g(x) when x = 1, we can substitute it into the expression for f(x) to find f(g(x)):

f(g(1)) = f(-1)

Substituting into the expression for f(x):

f(-1) = (-1)^2 + 3(-1) + 6
f(-1) = 1 - 3 + 6
f(-1) = 4

Therefore, f(g(x)) when x = 1 is 4. So, the correct answer is D) 4.