Simple “Apply” function

Question

This is such a simple question its almost embarrassing, and I'm surprised I haven't figured out where I'm going wrong with this..Probably because I'm really new to the commands Function and Apply and I might be using them wrong. But anyways, here it goes:

Input:

Clear[g2, x];
g2 := Function[x, Apply[Log[1 + x^2] + 6 x + x^3,x]]
g2[{2, 3}]

Output:

{20 + Log[5], 45 + Log[10]}[2, 3]

I have the answer but whats going on with the [2,3]?? I thought I need the extra "x" because Apply takes 2 arguments..

Here is an example I am trying to follow:

Input:

Clear[f2];
f2 := Function[pt, Sqrt[Apply[Plus, pt^2]]]
f2[{3, 4, 12}]

Output:

13

Answer 1

It is not clear what you are trying to do. The example you are trying to follow is not similar to your function g2. Since all of the functions used in g2 are Listable there is no need to use Apply

Clear[g2, x];
g2 := Function[x, Log[1 + x^2] + 6 x + x^3]
g2[{2, 3}]

(*  {20 + Log[5], 45 + Log[10]}  *)

Log[1 + #^2] + 6 # + #^3 &[{2, 3}]

(*  {20 + Log[5], 45 + Log[10]}  *)

If the components of g2 weren't Listable you could use Map

Log[1 + #^2] + 6 # + #^3 & /@ {2, 3}

(*  {20 + Log[5], 45 + Log[10]}  *)

Showing that they are all equivalent

% == %% == %%%

(*  True  *)