i have one big question that i want to know if i solved right, if i did wrong, please correct me and show me the right way to solve so i'll be able to learn.
Let f, g, h functions from R -> R prove or disprove: 1) if f * g = f * h, then g=h. my answer no, because even if the images equal there might be other values in g or h that might be not defined if f.
2)if f * g = g * h and f is injective(one-to-one) then g=h. here i said it's right because it's well defined and i don't think neither h or g might have values that are not in f(g) or f(h).
3)if g * f = h * f and f is supreme then g=h (function A -> B is called supreme if we can obtain B as a result of the function, wasn't sure if its name). Here i am not really sure, but i think it's true because it's well defined you can get it's image from only one member of the domain, so i think it's true.
4)if fg is rising, and f is going down so g is going down. i think it's true, because if we mark it using f(g(t), then if it's rising it's only if both are positive or both negative ((-t)(-t)).
5)if f * g i rising, and f is injective, then g monotonous. Here i am not sure, but i think it's false since being injective doesn't say a lot about f(g(x) to help us deduct that g is monotonous.
thank you very much for your help. hoping to learn from your answer and to understand this concept. if you can elaborate or correct me, it'll be very helpful. plus, i am sure others might profit from this.
note:by * i mean function combining, so f * g = f(g(x)).
Your answer of 1) is correct.
But for 2), it isn't. Assume that f(x)=x^2 and g(x)=x+1. Then f(g(x))=(x+1)^2, but g(f(x))=x^2+1, which are not the same.
Your answer of 3), 4) and 5) are probably true, but for 4), the word both should refer to the first derivative, not the original function.