Question

`AA x,y,z in RR f(x+y+z) = f(x)f(y)f(z)!=0` & `f(2)=5 , f'(0)=2` then find the value of `f'(2)` ?

Answer 1

`f'(2)=pm10`

Explanation:

`f(x+0+0)=f(x)f(0)^2=f(x)->f(0)^2= 1`
`f(x+delta+0)=f(x)f(delta)f(0)`

`lim_(delta->0)(f(x+delta)-f(x-delta))/(2delta) = lim_(delta->0)(f(x)f(delta)f(0)-f(x)f(-delta)f(0))/(2delta) =`
`= f(x)f(0)lim_(delta->0)(f(delta)-f(-delta))/(2delta) = f(x)f(0)f'(0)`

but `f(0)=pm1` and `f'(0) =2` so

`f'(x) = f(0)f'(0)f(x) = pm2f(x)` and finally

`f'(2)=pm10`

Note:

`(f'(x))/f(x)=pm1/2` then `f(x)=C_0e^(pm x/2)`