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Question Number 158687 by mahdipoor last updated on 07/Nov/21

\forall x \in R f (x) = f^{'} (x) a n d f (0) = 1

p r o v e f (a + b) = f (a) \times f (b)

Answered by mr W last updated on 07/Nov/21

y^{'} = \frac{d y}{d x} = y

\frac{d y}{y} = d x

\int \frac{d y}{y} = \int d x

\ln y - \ln C = x

\ln \frac{y}{C} = x

\Rightarrow y = {C e}^{x}

y (0) = C = 1

\Rightarrow y = f (x) = e^{x}

f (a + b) = e^{a + b} = e^{a} e^{b} = f (a) f (b)

Commented by mahdipoor last updated on 07/Nov/21

c a n y o u p r o v e w i t h o u t u s e e^{x} ?

Commented by mr W last updated on 07/Nov/21

f (x) = e^{x} i s t h e o n l y s o l u t i o n s a t i f y i n g

f^{'} (x) = f (x) a n d f (0) = 1 .

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