Let f(x) = ax^2 + bx + c where a,b,c are real numbers and a not equal to 0.
Show that if f[f(x)] = [f(x)]^2 for all x, then f(x) = x^2
AL pure math ~ polynomials
2009-01-11 5:31 am
回答 (2)
2009-01-11 6:32 am
2009-01-12 3:54 am
I provide another method
f(x)=ax2+bx+c
f'(x)=2ax+b
f''(x)=2a
The following steps(in red colour)are in order to find a
As f[f(x)] = [f(x)]2
Differentiating both sides w.r.t. x
f'[f(x)]f'(x)=2f(x)f'(x)
f'[f(x)]=2f(x)--------------------------(1) (as a not equal to 0,f'(x) also not equal to 0,so f'(x) can be eliminated in both sides)
Differentiating both sides w.r.t. x
f''[f(x)]f(x)=2f'(x)
f''[f(x)]=2
2a=2
a=1
The following steps(in blue colour)are in order to find b
From (1)
f'[f(x)]=2f(x)
2af(x)+b=2f(x) , as f'(x)=2ax+b , so f'[f(x)]=2af(x)+b
2(1)f(x)+b=2f(x)
b=0
The following steps(in green colour)are in order to find c
f(x)=x2+c
f[f(x)] = [f(x)]2
Put x=0
f[f(0)]=[f(0)]2
f(c)=c2
c2+c=c2
c=0
so if f[f(x)] = [f(x)]2 for all x, then f(x) = x2
f(x)=ax2+bx+c
f'(x)=2ax+b
f''(x)=2a
The following steps(in red colour)are in order to find a
As f[f(x)] = [f(x)]2
Differentiating both sides w.r.t. x
f'[f(x)]f'(x)=2f(x)f'(x)
f'[f(x)]=2f(x)--------------------------(1) (as a not equal to 0,f'(x) also not equal to 0,so f'(x) can be eliminated in both sides)
Differentiating both sides w.r.t. x
f''[f(x)]f(x)=2f'(x)
f''[f(x)]=2
2a=2
a=1
The following steps(in blue colour)are in order to find b
From (1)
f'[f(x)]=2f(x)
2af(x)+b=2f(x) , as f'(x)=2ax+b , so f'[f(x)]=2af(x)+b
2(1)f(x)+b=2f(x)
b=0
The following steps(in green colour)are in order to find c
f(x)=x2+c
f[f(x)] = [f(x)]2
Put x=0
f[f(0)]=[f(0)]2
f(c)=c2
c2+c=c2
c=0
so if f[f(x)] = [f(x)]2 for all x, then f(x) = x2
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