#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)# ?

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#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)# ?

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Answer 1 · 2016-10-13T17:03:59

#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)#

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#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)# ?

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