Answer:
Only in special funcitons, specifically linear functions whose graphs pass through the origin, it is the same to put the constant inside the argument or outside
Step-by-step explanation:
If the function has the form f(x) = a*x, where a is a constant, then we have that f(c*x) = a*(c*x) = c*(a*x) = c*f(x). This kind of functions are proportional to the identity function f(x) = x, and they comprehend the linear functions whose graph pass through the origin.
For other linear function this property isnt true. For example if f(x) = x+4, then
f(4) = 4+4 = 8,
f(2*4) = 8+4 = 12
2*f(4) = 2*8 = 16
Thus, 2*f(4) is not f(2*4).
This property also isnt true for quadratic functions for example. If f(x) = x², then f(1) = 1, thus 3*f(1) = 3, however, f(3*1) = 3² = 9.
There might be coincidences for specific values, for example if f(x) = (x-1)*(x-2), then f(2*1)=2*f(1) = 0, however, for any other constant the result is not the same (for example f(0*1) = (-1)*(-2)=2, and 0*f(1) = 0).
If we want a function to satisfy the property f(c*x) = c*f(x) for any c,x, then it should be true that f(c) = f(c*1) = c*f(1). This means that if f(1) = a, then f(c) = c*a, so, in other words, f(x) = a*x = f(1) * x.
