
Arturo O. answered 06/03/20
Experienced Physics Teacher for Physics Tutoring
None are true for EVERY single function f(x). I will give you true and false examples.
(1)
f(x) = kx,
where k is a constant.
f(a + b) = k(a + b) = ka + kb = f(a) + f(b)
Now consider
f(x) = x2
f(a + b) = (a + b)2 = a2 + 2ab + b2 ≠ f(a) + f(b)
(2)
f(z) = z2
f(ab) = (ab)2 = a2b2 = f(a)f(b)
Now consider
f(x) = x + 1
f(ab) = ab + 1
f(a)f(b) = (a + 1)(b + 1) = ab + a + b + 1 ≠ ab + 1 (unless a+b=0)
f(ab) ≠ f(a)f(b) in this case
Now, can you apply this kind of reasoning and show that it is not generally true that f(1/a) = 1/f(a)?

Arturo O.
Yes, you understood correctly. Regarding the grade where you learn this, it will probably be between the freshman and junior years of high school, depending on the school. What grade are you in now?06/03/20

Arturo O.
By the way, the last example in your Comment should end with 1/f(a) = 2/a.06/03/20
Minhui R.
Oh that's true. Actually, I'm preparing for Cegep in Canada but I've never seen this type of question where we compare two functions and determine if they are equal.06/03/20
Minhui R.
okay Let's say f(x)=x^(1/2) f(1/a)=(1/a)^(1/2)=1/a^(1/2) and 1/f(a)=1/a^(1/2) so in this case, f(1/a)=1/f(a) Now let's say f(x)=0.5x f(1/a)=0.5(1/a)=0.5/a and 1/f(a)=1/0.5a=2a in conclusion, f(1/a)≠1/f(a) here. Did I understand correctly? And may I ask which grade do we learn this stuff? Thank you in advance!06/03/20