兩次方程問題(20分)

For a=0
f(x)=bx+c=x is no real root.
(b-1)x=-c is no real root.
b=1,c is not equal 0.
f(f(x))=f(x)+c=x+2c=x
so that f(f(x))-x is no real root .

For a>0
f(x) no real root, f(x)-x>0
f(f(x))-f(x)>0
f(f(x))>f(x)>0

For a<0
f(x)-x<0
f(f(x))<f(x)<0

網上有答案，我再打一次轉載給你看，參考網址在下方。

Consider g(x) = f(x) - x.
Note that f is continuous, then g is continuous too.
f(x) = x has no real root, that is, g(x) = 0 has no real root.
g is continuous, that means (Case 1) g(x) always >0 or (Case 2) g(x) always <0.

Therefore, f(x) always >x or f(x) always < x.
This means, f(y)>y for all y or f(y)<y for all y.
Now, put y = f(x) for both cases.
Case 1:
f(f(x))>f(x) for all f(x) with f(x) always >x
Therefore, f(f(x))>f(x)>x for all x.
That is, f(f(x)) is NOT equal to x for all x.

Case 2:
f(f(x))<f(x) for all f(x) with f(x) always <x
Therefore, f(f(x))<f(x)<x for all x.
That is, f(f(x)) is NOT equal to x for all x.

Combining, f(f(x)) can never be x.
That is, f(f(x))=x has no real roots.