Prove that there is the unique function f(x), such as f'(x)=4f(x), if x is real and f(0)=4.

We'll create a function h(x), such as h(x) =
f(x)*e^(-4x)

This function is differentiable, with respect
to x, over the real set of numbers. Since h(x) is a product of functions, we'll apply
product rule, to differentiate it:

(u*v)' = u'*v +
u*v'

h'(x) = f'(x)*e^(-4x) -
4e^(-4x)*f(x)

We'll factorize by
e^(-4x):

h'(x) = e^(-4x)*[f'(x) -
4*f(x)]

Since e^(-4x)>0, only the factor [f'(x) -
4*f(x)] can be cancelled.

Since h'(x) = [f'(x) - 4*f(x)] =
0 => h(x) = k => f(x)*e^(-4x) = k

But, from
enunciation, we know that f(0) = 4:

f(0)*e^(0) =
k

4*1 = k => k =
4

Finally, f(x) =
4*e^(4x)

The requested function, taht
respects all the given constraints, is f(x) =
4*e^(4x).