Let $f(x+y)=f(x)f(y)$ for all $x$ and $y$ and... 
Let $f(x+y) = f(x)f(y)$ for all $x$ and $y$, and $f(5) = -2$, $f'(0) = 3$, then find the value of $f'(5)$.

I thought a lot on this question. I tried to find out the function. I put $y=0$ to get $f(0)$ which comes out to be $1$. but then I could do no more.
 A: We have $f(0)=1$ or $f(0)=0$. If $f(0)=0$ , then $f(x)=0$ for all $x$, which is not the case. Hence $f(0)=1$.
$f'(5)= \lim _{h \to 0}\frac{f(5+h)-f(5)}{h}=f(5) \lim _{h \to 0}\frac{f(h)-1}{h}=-2\lim _{h \to 0}\frac{f(h)-f(0)}{h}=-2 f'(0)=-6$
A: There is a problem with this problem:  It cannot have a solution, because the hypotheses lead to a contradiction.
As others have noted,
$$f'(x)=\lim_{h\to0}{f(x+h)-f(x+0)\over h}=\lim_{h\to0}{f(x)f(h)-f(x)f(0)\over h}=f(x)\lim_{h\to0}{f(0+h)-f(0)\over h}=f(x)f'(0)$$
so $f$ is differentiable at all points, which requires $f$ to be continuous at all points.  But we also have $-2=f(5)=f(5+0)=f(5)f(0)=-2f(0)$, which implies $f(0)=1$.  And now, since $f(0)\gt0$ while $f(5)\lt0$, the Intermediate Value Theorem implies $f(a)=0$ for some $a\in(0,5)$.  But that implies $f(x+a)=f(x)f(a)=0$ for all $x$, and that's a contradiction.
A: Take the equation $f(x)f(y) = f(x+y)$.
Now, on both sides, take the derivative with respect to $x$ (i.e., $y$ is a constant, and so is $f(y)$.
You get $$f'(x) f(y) = f'(x+y)$$
Now set $x=5, y=-5$ and you get$$f'(5)=\frac{f'(0)}{f(-5)}$$
So all you need to calculate $f'(5)$ is to know what $f(-5)$ is. To calculate that, set $x=5,y=-5$ in $f(x+y)=f(x)f(y)$ and you get $$f(0)=f(5)f(-5)$$

Edit thanks to @BarryCipra:
A shorter way to do it is to plug in $x=0, y=5$ and get$$f'(0)f(5)=f'(5)$$ immediatelly.
A: The question itself as presented now is invalid.
By taking $y=x$ we get $f(2 x) = f(x)^2$ from which it follows by induction that $f(n x) = f(x)^n$ for any integer $n$ and any value of $x$.
Taking $x=\frac{1}{2}$ and $n=10$ we get $f(\frac{1}{2})^{10}=-2$ and hence $f(\frac{1}{2})$ needs to be a complex variable and hence the function $f(x)$ is complex. Note, that I did not use any derivatives yet.
This is not necessarily a problem in itself, but already looks suspicious for this type of problem.
We can also take the limit $\delta \rightarrow 0$ in $f(n \delta) = f(\delta)^n$ and since the question implies the existence of the derivatives locally the function $f(x)$ would have to be some exponential $f(x)=e^{\alpha x}$ for some complex constant $\alpha$. If we use the condition $f'(0)=3$ this gives $f(x)=e^{3x}$ but than the condition $f(5)=-2$ can not be satisfied.
The main issue here is the fact that the requirement $f(x+y)=f(x) f(y)$ is supposedly valid for all $x,y \in \mathbb{R}$. If one would limit this range, for instance by restricting $x,y \in [-1,1] \cup [-4,6]$ the problem could be solved by creating some discontinuity in the function $f(x)$in a region not covered by the requirement.
A: A very simple argument that the problem is invalid:
We have $f(2x)=f(x+x)=f(x)^2 \ge 0$ for all $x$. Hence $f(t) \ge 0$ for all $t$. But then $f(5) = -2$ is impossible !
