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Given: f(x+y)=f(x)f(y). f'(0) exists.

Show that f is differentiable on R.

At first, I tried to somehow apply the Mean Value Theorem where f(b)-f(a)=f'(c)(b-a). I ended up lost...

Then I tried showing f(0)=1, because f(x-0)=f(x)f(0) and f(x) isn't equal to 0.

However, with that conclusion, f'(0)=0. This is a problem, because the function is most likely f(x)=e^cx, and f'(x)=ce^cx, and f'(0)=c, which may not be 0....

Am I making this problem harder than it is? Any help would be greatly appreciated!

http://eqworld.ipmnet.ru/en/solutions/fe/fe4101.pdf

Show that f is differentiable on R.

At first, I tried to somehow apply the Mean Value Theorem where f(b)-f(a)=f'(c)(b-a). I ended up lost...

Then I tried showing f(0)=1, because f(x-0)=f(x)f(0) and f(x) isn't equal to 0.

However, with that conclusion, f'(0)=0. This is a problem, because the function is most likely f(x)=e^cx, and f'(x)=ce^cx, and f'(0)=c, which may not be 0....

Am I making this problem harder than it is? Any help would be greatly appreciated!

http://eqworld.ipmnet.ru/en/solutions/fe/fe4101.pdf

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