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function outside(x) 
 function inside(y) 
 return x + y;
 
 return inside;

fn_inside = outside(3); //1* - Think of it like: give me a function that adds 3 to whatever you give it
result = fn_inside(5); //2* - returns 8

result1 = outside(3)(5); // returns 8

1*,2* - Since each call provides potentially different arguments, a new closure is created for each call to outside. The memory can be freed only when the returned inside is no longer accessible.

Questions

1) What happens when we call outside(3)?

• 1a) What does it return?

2) Is fn_inside(3); a valid function call?

3) How is outside(3)(5) valid when outside function definition
accepts only one parameter?

1. 
   outside(3) returns a function. Like you said, it returns a function takes one argument n and returns n + 3. Mathematically, this might be written as f(n) = n + 3.


2. 
   fn_inside is a binding made to the return value of outside(3), which we know, from part 1, is a function. So when we call fn_inside(3), it's like plugging in 3 for n: f(3) = 3 + 3 = 6. The return value is simply a number: 6.


3. 
   outside(3)(5) is equivalent to (outside(3))(5) and recall that outside(3) is a function that takes one argument.