The function y is defined on r = f (x), and f (0) is not equal to 0. When x >; 0.f(x)> 1, and f (x+y) = f (x) × f (y) for any real number x, y/. 0. If there is 0 1, find the value range of x.

Solution: Let x=y=0, then f(0)=f(0)f(0).

∫f(0)≠0

∴f(0)= 1

Let y=-x, then f(x-x)=f(x)(-x)= 1.

∴f(-x)= 1/f(x)

∵x & gt； 0.f(x)> 1

∴-x<； 0

∴0<； f(x)& lt； 1

That is, when x

(2) make X 1

f(x 1)-f(x2)= f(x 1)-f(x2-x 1+x 1)

= f(x 1)-f(x2-x 1)f(x 1)

= f(x 1)( 1-f(x2-x 1))

∵x2-x 1 & gt； 0

∴f(x2-x 1)>； 1

∴f(x 1)<； f(x2)

It's adding functions.

(3)f(x^2)*f(2x-x^2+2)>； 1

f(x^2+2x-x^2+2)>； f(0)

It's adding functions.

∴x^2+2x-x^2+2>； 0

x & gt- 1