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posted image:*

Exact replication of original problem statement as given in the posted image:

$\text{Let }\,f(x+y) = f(x)\cdot f(x)\quad \forall x, y \in \mathbb R.$

Suppose $f(5) = 10$ and $f'(0) = 5$.

Then the value of $f'(5)$ is

$(a)\quad 5$
$(b) \quad 10$
$(c) \quad 100$
$(d) \quad$None of the above.

How to solve this type of question? please explain ...


EDIT
There may be a typo in the given problem statement. So any hints on solving the question $$f(x + y) = f(x) \cdot f(y)\;$$

given $f(5) = 10$ and $f'(0) = 5$.

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closed as off-topic by Rory Daulton, Jack D'Aurizio, user147263, Carl Mummert, Did Feb 22 '15 at 7:41

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If this question can be reworded to fit the rules in the help center, please edit the question.

  • 2
    $\begingroup$ What does $f(x).f(x)$ mean? Is that multiplication? $\endgroup$ – Rory Daulton Feb 21 '15 at 16:15
  • $\begingroup$ @ Rory Daulton... its multiplication... $\endgroup$ – user45799 Feb 21 '15 at 16:19
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    $\begingroup$ And I suppose it's f(x).f(y) $\endgroup$ – Tryss Feb 21 '15 at 16:23
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    $\begingroup$ See my edit, Prasanta. $\endgroup$ – Namaste Feb 21 '15 at 16:31
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    $\begingroup$ the idea is to use the additive rule $f(x+y) = f(x)f(y)$ of $f$ to derive the fact $f'(5) = f'(0)f(5).$ in fact $5$ is not special and in general $f'(x)=f'(0)f(x).$ $\endgroup$ – abel Feb 21 '15 at 16:45
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Assuming a typo in the exercice and using $f(x+y)=f(x)f(y)$. And correct regularity

The idea is to use the definition of the derivative :

$$f'(a) = \lim_{h\to 0} \frac{f(a+h)-f(a)}{h}$$

Using the propriety of f, we get :

$$f'(5) = \lim_{h\to 0} \frac{f(5+h)-f(5)}{h} =\lim_{h\to 0} \frac{f(5)f(h)-f(5)}{h} =\lim_{h\to 0} f(5) \frac{f(h)-1}{h} $$

Now, we get that $f(0) = 1$, because $f(5) = f(5+0) = f(5)f(0)$, but $f(5)\neq 0,$ then $f(0) = 1$

$$ f(5) \lim_{h\to 0} \frac{f(h)-1}{h} = f(5) \lim_{h\to 0} \frac{f(0+h)-f(0)}{h} = f(5) f'(0) = 50$$

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  • $\begingroup$ Perfect solution. $\endgroup$ – Arashium Feb 21 '15 at 16:34
  • $\begingroup$ Not perfect, it lack an argument why f(0) can't be 0... will modify $\endgroup$ – Tryss Feb 21 '15 at 16:35
  • $\begingroup$ if $f(0)=0$ then $f(x)=0$ for all $x$ $\endgroup$ – Arashium Feb 21 '15 at 16:37
  • $\begingroup$ @Tryss: you can actually use $\lim$ - it kinda looks nicer $\endgroup$ – Alex Feb 21 '15 at 20:02
  • $\begingroup$ You can also get $f(0)=1$ from $f(2x) = f^2(x)$ and then differentiate and set $x=0$. $\endgroup$ – Alex Feb 21 '15 at 20:03
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If the equation is

$$f(x+y)=f(x)\cdot f(x)$$

then if we let $y=0$ we get

$$f(x)=[f(x)]^2$$

This means that $f(x)=0$ or $f(x)=1$. (Check the solution to $y=y^2$.)

If $f(x)$ is constantly $0$ or $1$ in an interval around any $x$, the derivative will be zero. If not, the derivative will be undefined. Either way, it is impossible for $f'(0)=5$ so the problem is badly posed.


Now let's assume the correct equation is

$$f(x+y)=f(x)\cdot f(y)$$

Let's also assume that $f(x)$ is continuous on all real numbers.

If we let $x=5,\ y=0$ we get

$$f(5)=f(5)\cdot f(0)$$

Then either $f(5)=0$ or $f(0)=1$. We are told $f(5)=10$ so we must have $f(0)=1$.

We also see that

$$f(2x)=f(x+x)=f(x)\cdot f(x)=[f(x)]^2$$ $$f(3x)=f(2x+x)=f(2x)\cdot f(x)=[f(x)]^3$$

By induction we can prove

$$f(nx)=[f(x)]^2$$

for integral $n$ .

You can also prove this for rational $n$, and continuity makes it true for real $n$. So, we see that

$$f(x)=e^{kx}$$

for some constant $k$.

Can you continue from there?

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  • $\begingroup$ Seems there is a mistake in the question, Maybe they mean $f(x+y)=f(x)f(y)$ $\endgroup$ – Arashium Feb 21 '15 at 16:25
  • $\begingroup$ @ Arashium...how to solve if it is f(x).f(y). $\endgroup$ – user45799 Feb 21 '15 at 16:26
  • $\begingroup$ @Prasanta, it needs to know $f'(10)$ to solve it. $\endgroup$ – Arashium Feb 21 '15 at 16:29
  • $\begingroup$ @ Arashium...please explain... $\endgroup$ – user45799 Feb 21 '15 at 16:31
  • $\begingroup$ @Prasanta $$f(2x)=f^2(x)$$ $$2f'(2x)=2f(x)f'(x)$$ $$f'(2x)=f(x)f'(x)$$ $$f'(10)=f(5)f'(5)$$ $\endgroup$ – Arashium Feb 21 '15 at 16:32

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