1. if a function is continuous and differentiable every where and satisfies the following property f(0)=1=f ' (0)and f(a+b)= f(a)f(b); a, b constants then prove f' '(x)=f(x) can you give an example of such a function?
please steps by steps !
1. by definition, lim(h -> 0) [f(h)-f(0)]/h= f'(0) =1
2. f'(x)= lim(h -> 0) [f(x+h)-f(x)]/h
=lim(h -> 0) [f(x)*f(h) - f(x)]/h
=f(x)* lim(h -> 0) [f(h)-1]/h
=f(x)* 1=f(x)
so, f'(x)=f(x)
3. f"(x)=[f'(x)]'=f'(x)=f(x)
4. For instance, f(x)=e^x
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Verified answer
1. by definition, lim(h -> 0) [f(h)-f(0)]/h= f'(0) =1
2. f'(x)= lim(h -> 0) [f(x+h)-f(x)]/h
=lim(h -> 0) [f(x)*f(h) - f(x)]/h
=f(x)* lim(h -> 0) [f(h)-1]/h
=f(x)* 1=f(x)
so, f'(x)=f(x)
3. f"(x)=[f'(x)]'=f'(x)=f(x)
4. For instance, f(x)=e^x