Question

Let a function $f:\Re \to \Re$ be given by $f(x + y) = f(x)f(y)$ for all$x,y \in \Re$. If the function$f(x)$ is differentiable at$x = 0$, to show that $f'(x) = f'(0)f(x)$ for all$x \in \Re$and also to determine the value of$f(x)$.

Answer 1

we have already known that If$f:\Re \to \Re$ is a function
Then, $(f(x)f(y))' = f'(x)f(y) + f(x)f'(y)$Where, $f(x)$ is differentiable on$\Re$.

Complete step-by-step answer:
Let us consider the given function $f:\Re \to \Re$ which is given by $f(x + y) = f(x)f(y)$ for all $x,y \in \Re$.
It is also given that the function$f(x)$ is differentiable at $x = 0$
Here we have that is, $f(x + y) = f(x)f(y)$ in given.
Now we are going to differentiate the given function for$x$, we get,
$f'(x + y) = (f(x)f(y))'$
From the given hint we can substitute the value of$(f(x)f(y))'$,
Hence now we get after the substitution of the value of$(f(x)f(y))'$,
$f'(x + y) = f'(x)f(y) + f(x)f'(y)$
We have to find the value of the function $x = 0$, we will substitute the $x$by zero.
So substitute $x = 0$ in the function we get,
$f'(0 + y) = f'(0)f(y) + f(0)f'(y)$
Here, $f(y)$is a function of$y$. So, $f'(y) = 0$ for any values of$x$.
Now we are going to substitute $f'(y) = 0$ in the above-given step we get,
$f'(y) = f'(0)f(y)$
Since $y$ is a variable, we can change the variable $y$ by a new variable $x$ since$x,y \in \Re$.
If the function$f(x)$ is differentiable at$x = 0$,
$f'(x) = f'(0)f(x)$for all $x \in \Re$
Hence we have shown that$f'(x) = f'(0)f(x)$for all $x \in \Re$.
As we have proved the above function we know that,
$f'(x) = f'(0)f(x)$
Now we have to find$f(x)$ of the function we can divide the above equation by$f'(0)$on both sides,
On dividing the term we get,
$f(x) = \dfrac{{f'(x)}}{{f'(0)}}$, provided $f'(0) \ne 0$
Now we have determined that $f(x) = \dfrac{{f'(x)}}{{f'(0)}}$for all $x \in \Re$.
Hence, we have shown that $f'(x) = f'(0)f(x)$ for all $x \in \Re$
And,

$f(x) = \dfrac{{f'(x)}}{{f'(0)}}$, for all $x \in \Re$ provided $f'(0) \ne 0$

Note:
If $f'(0) = 0$then the determined functional value of $f(x)$ will become undefined. Hence not to get an undefined value for$f(x)$the differentiation of the function at $x = 0$has to be a non-zero value.