$AA x,y,z in RR f(x+y+z) = f(x)f(y)f(z)!=0$ & $f(2)=5 , f'(0)=2$ 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 $f'(2)$ ?

1 Answer

Explanation:

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Math

Humanities

... and beyond

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

1 Answer

Explanation:

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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 $f'(2)$ ?