- #1
musamba
- 2
- 0
Hi.
I have a problem with prooving this:
Q: Assume a function [tex] f(x): R --> R [/tex] is a continuous function such as [tex] f(x + y) = f(x)f(y) [/tex].
Proove that [tex] f [/tex] is deriveable if [tex] f'(0) [/tex] exists.
I've done this so far:
[tex] f(x) = f(x + 0) = f(x)f(0)[/tex]
Then I use the product rule:
[tex]\\ f'(x) = f'(x)f(0) + f(x)f'(0)[/tex]
And manipulate the equation:
[tex]\\ \\ f'(x) - f'(x)f(0) = f(x)f'(0)[/tex]
[tex]\\ \\ f'(x)(1-f(0)) = f(x)f'(0)[/tex]
[tex]\\ \\ (*) f'(x) = \frac{f(x)f'(0)}{1-f(0)}[/tex]
But this must be wrong when:
[tex]\\ f(x) = f(x + 0) = f(x)f(0)[/tex]
Wrtiting different gives:
[tex]\\ f(x) = f(x)f(0)[/tex]
[tex]\\ \\ f(0) = \frac{f(x)}{f(x)}[/tex]
[tex]\\ \\ f(0) = 1[/tex]
When [tex] f(0) = 1 , (*)[/tex] only exist when [tex]f(0)[/tex] is not equal to 1.
Please tell me what went wrong in my calculation.
I have a problem with prooving this:
Q: Assume a function [tex] f(x): R --> R [/tex] is a continuous function such as [tex] f(x + y) = f(x)f(y) [/tex].
Proove that [tex] f [/tex] is deriveable if [tex] f'(0) [/tex] exists.
I've done this so far:
[tex] f(x) = f(x + 0) = f(x)f(0)[/tex]
Then I use the product rule:
[tex]\\ f'(x) = f'(x)f(0) + f(x)f'(0)[/tex]
And manipulate the equation:
[tex]\\ \\ f'(x) - f'(x)f(0) = f(x)f'(0)[/tex]
[tex]\\ \\ f'(x)(1-f(0)) = f(x)f'(0)[/tex]
[tex]\\ \\ (*) f'(x) = \frac{f(x)f'(0)}{1-f(0)}[/tex]
But this must be wrong when:
[tex]\\ f(x) = f(x + 0) = f(x)f(0)[/tex]
Wrtiting different gives:
[tex]\\ f(x) = f(x)f(0)[/tex]
[tex]\\ \\ f(0) = \frac{f(x)}{f(x)}[/tex]
[tex]\\ \\ f(0) = 1[/tex]
When [tex] f(0) = 1 , (*)[/tex] only exist when [tex]f(0)[/tex] is not equal to 1.
Please tell me what went wrong in my calculation.
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