Page 28 - Buku Siap OSN Matematika SMP 2015(1)

Aljabar

1 = f(–1)  f(–1) = 1

f(xy) = f(x – y)

f(–11) = f(–1 – 1)

f(–1) = f(–2)

1 = f(–2)  f(–2) = 1

Jadi, f(–2) – f(4) = 1 – 1 = 0.

2x  4
x
4. f ( )  , x  0 dan x bilangan real, maka f 2009 (6) = ....
x
2
3
Catatan: Notasi f (x) = f ( f(x)), notasi f (x) = f( f( f(x))) , dan seterusnya.
Jawab:


f (6)  2 6 4  4
6 3

4
2  4
6
f 2 (6)   f f    3   1
4
3
2   1  4
6
f 3 (6)    f f f     6
  1


6
f 4 (6)   f f 3    2 6 4  4
6 3
4
2  4
6
f 5 (6)   f f 4    3   1
4
3

2   1  4
f 6 (6)   f f 5     6
6
  1


Siap OSN Matematika SMP 2015 19