If x is greater than 0, f (x) is greater than 1, f (a + b) = f (a). F (b) It is proved that FX is a proper function on R, a and B belong to R Increasing function

If x is greater than 0, f (x) is greater than 1, f (a + b) = f (a). F (b) It is proved that FX is a proper function on R, a and B belong to R Increasing function

f(0)=f(0)*f(0),
f(0)=1
Let X1 > X2, x1-x2 > 0,
f(x-x)=f(0)=f(x)f(-x)=1
So f (- x) = 1 / F (x)
Because x belongs to [0, positive infinity) f (x) > 1,
So when x belongs to (negative infinity, 0), - x belongs to [0, positive infinity],
f(x)=1/f(-x)>0
f(x1-x2)=f(x1)f(-x2)=f(x1)/f(x2)>1
The results show that f (x1) > F (x2)
So it's an increasing function