Given that f(x) = x2 + 3x + 6 and g(x) = the quantity of x minus three, over two, solve for f(g(x)) when x = 1.
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x^2=(-1)^2=1 +3x=3(-1)=-3 =+6 adding these 1-3+6=4
Basically, lets say f(x) was 1+x and g(x) was 2+x, when x = 3. f(g(x)) basically says to plug g(x) where ever you see an x in f(x), so you would do 1+(2+x). You see it?
Next, you would plug in 3 to find the answer, which would be 6.
Same with your problem. f(x) is x^2+3x+6 and g(x) = (x-3)/2
So where you see a x in f(x), put the equation of g(x)
y = ((x-3)/2)^2 + (3(x-3)/2) + 6
Which is completely bleh. You would then plug 1, which was x=1 the given, into that equation for your answer.
y = ((1-3)/2)^2 + (3(1-3)/2) + 6
y = 1+(-3)+6
y = 4