Derivate Question: Proving f'(0) Exists

  • Thread starter musamba
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In summary, the problem is that you're trying to use the product rule to show that the derivative of f(x+y) exists, when only the exponential functions satisfy that condition.
  • #1
musamba
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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.
 
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  • #2
musamba said:
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]
This is your error. f'(0) means f'(x) evaluated at x=0, not "the derivative of f(0)". f(0) is a constant and necessarily has derivative 0. You are just saying that f'(x)= f'(x)f(0). (And so showing in a different way that f(0)= 1.)

I am also not happy with using the product rule to show that f ' does exist. Part of the hypotheses for the product rule is that the derivatives must exist.
Go back to the original definition:
[tex]f'(x)= lim_{h \rightarrow 0}\frac{f(x+h)- f(x)}{h}[/tex]
using the fact that f(x+h)= f(x)f(h), that f '(0) exists, and that f(0)= 1.\

Your proof of that:
[tex] f(x) = f(x)f(0)[/tex]
[tex]f(0) = \frac{f(x)}{f(x)}[/tex]
[tex]f(0) = 1[/tex]
is correct.

It can be shown, by the way, that the only continuous (and so only differentiable) functions satisfying f(x+y)= f(x)f(y) are the exponential functions: f(x)= ax.
 
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  • #3
Thank you, HallsofIvy.
I will give that a shot.
 

1. What is a derivative?

A derivative is a mathematical concept that represents the rate of change of a function at a specific point. It is the slope of a tangent line at that point and provides information about the behavior of the function.

2. How do you prove that f'(0) exists?

To prove that f'(0) exists, you need to show that the limit of the difference quotient as x approaches 0 exists. This means that the left and right-hand limits must be equal, indicating that the derivative at x=0 is well-defined.

3. Why is proving f'(0) exists important?

Proving that f'(0) exists is important because it shows that the function is differentiable at x=0. This means that the function has a well-defined derivative at that point, which is necessary for many applications in mathematics and science.

4. Can f'(0) exist for a function that is not continuous at x=0?

No, f'(0) cannot exist for a function that is not continuous at x=0. Since continuity is a necessary condition for differentiability, a function must be continuous at x=0 in order for f'(0) to exist.

5. How is the existence of f'(0) related to the graph of the function?

The existence of f'(0) is related to the graph of the function by the fact that the function must have a well-defined tangent line at x=0 in order for f'(0) to exist. This means that the graph of the function must be smooth and have no sharp turns or breaks at x=0.

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