Question 515560
When you write (f o g)(x), you mean the composition of f after g, or f(g(x)) (notice that f is to the left of g, just like in (f o g)(x)).  In other words, first you plug x into g(x), then you plug the result from g(x) into the function f.  In this case, we have

{{{f(x) = 2x^2 + 3}}} and {{{g(x) = x - 5}}}

and we are asked to compute (f o g)(3) and (g o f)(3).

For (f o g)(3), you are first plugging 3 into g(x):

{{{g(3) = 3 - 5 = -2}}}

Now you take g(3) = -2 and plug that result into f:

{{{f(g(3)) = f(-2) = 2*((-2)^2) + 3 = 2 * 4 + 3 = 8 + 3 = 11}}}

For (g o f)(3), the situation is reversed: you are computing g(f(3)) (notice how the order changed), so you first plug 3 into f(x):

{{{f(3) = 2 * (3^2) + 3 = 2 * 9 + 3 = 18 + 3 = 21}}}

and then take f(3) = 21 and plug that result into g:

{{{g(f(3)) = g(21) = 21 - 5 = 16}}}

So (f o g)(3) = 11, while (g o f)(3) = 16.  Now, we could have gotten the same answer by plugging the actual formula for g(x) into f(x) and vice versa, like you started to do, and you may be asked to do that on an exam (you have the correct formula for (f o g)(x)).  That said, if you just need to compute the composition at some value of x, like x = 3, using the method above is usually the easiest way to go.