Let $f(x)$ be a real valued, differentiable function such that for any $x,y \in \mathbb{R}$,$f(x+y)=f(x)f(y)$. Suppose there exist $a,b$ such that $f(a)\neq 0, f'(b)>0$.
Show that $f(x)$ is a monotonic function
I tried to use the Cauchy equation result to get: $f(x)=p^x$ for some $p>1$.
I would like to know if there are simpler methods.
Note that $\frac{f(x+h)-f(x)}{h}=f(x)\frac{f(h)-f(0)}{h}$, so $f'(x) = f(x)f'(0)$, and that $f$ is a positive function.
Putting these two together gives $f'(b) = f(b)f'(0)>0$, so $f'(0)>0$. Thus for any $x$, $f'(x) \geq 0$ so the function is monotone increasing.
Differentiate your equation w.r.t. $x$ and (after that) set $ x=0$. This gives $f'(y) = f'(0) f(y)$ for all $y$ and thus $f(y) = f(0) e^{f'(0) y}$ or all $y$. This should allow you to complete the proof.
EDIT: Even without ODE, consider
$$ \frac{d}{dx} \frac{f(x)}{e^{f'(0)x}} = \frac{f'(x) e^{f'(0)x} - f(x) f'(0) e^{f'(0)x}}{e^{2f'(0)x}} = 0, $$
which implies the above formula for $f(y)$.
Watch out: Is there a name for function with the exponential property $f(x+y)=f(x) \cdot f(y)$?
Now assuming $f(x)$ is differentiable and continious we have that only $f(x) = e^{cx}$ statisfy the equation, but $e^{cx}$ strictly increasing and hence strictly monotone.
The conditions $f(x+y)=f(x)f(y)$ and $f(a)\neq 0$ imply that $f(x)\neq 0$ for all $x\in\mathbb R$. Since $f$ is continuous, you can assume that $f$ is strictly positive (or consider $-f$ if $f$ is negative instead). Consider then the function $g(x)=\ln(f(x))$, which satisfies $$g(x+y)=g(x)+g(y),\;\;g'(x)=f'(x)/f(x).$$ Now, $g'(x+y)=\partial_x g(x+y)=g'(x)$, so $$0<f'(b)/f(b)=g'(b)=g'(b+(x-b))=g'(x)$$ for all $x\in\mathbb R$. In particular, $g$ is strictly increasing, and so is $f=\exp\circ g$.
I will give it a try - without the idea that $f(x) = a \exp(x)$...
Let $f(x)$ be a real valued, differentiable function such that for any $x,y \in \mathbb{R} ,f(x+y)=f(x)f(y)$. Suppose there exist $a,b$ such that $f(a) \ne 0$ ,$f'(b) > 0$.
Show that $f(x)$ is a monotonic function
We have
$$\forall\ x \in \mathbb{R} : f(x) \in \mathbb{R},\tag{1}$$ $$\forall\ x,y \in \mathbb{R} : f(x+y) = f(x) f(y),\tag{2}$$ $$\exists\ a \in \mathbb{R} : f(a) \ne 0,\tag{3}$$ $$\exists\ b \in \mathbb{R} : f'(b) > 0.\tag{4}$$
First
$$ f(x) = f\left( \frac{x}{2} + \frac{x}{2} \right) = f^2\left( \frac{x}{2} \right) \ge 0.\tag{5} $$
Second
$$\Big\{ \exists\ y: f(y) = 0 \Rightarrow \forall\ x: f(x) = f(x-y) f(y) = 0 \Big\} \tag{6} $$
$$\Downarrow$$
$$\Big\{ \exists\ y: f(y) \ne 0 \Rightarrow \forall\ x: f(x) \ne 0 \Big\},\tag{7} $$
whence $f(x) \ne 0$, because $f(a) \ne 0$. Combination of $(5)$ and $(7)$ yields
$$ f(x) > 0.\tag{8} $$
Third
$$ f'(x) = \lim_{y \rightarrow 0} \frac{ f(x + \color{red}{y}) - f(x + \color{blue}{0})}{y} = f(x) \lim_{y \rightarrow 0} \frac{ f(\color{red}{y}) - f(\color{blue}{0})}{y} = f(x) f'(0),\tag{9} $$
whence
$$ f'(0) = \frac{f'(x)}{f(x)}.\tag{10} $$
As $f'(b) > 0$ and $f(b) > 0$ from $(8)$, we obtain
$$f'(0) > 0.\tag{11}$$
Fourth
As $f(x) > 0 $ from $(8)$ and $f'(0) > 0$ from $(11)$, we obtain
$$ f'(x) > 0,\tag{12} $$
from $(9)$.
Conclusion:
As $f'(x) > 0$, the function $f(x)$ is monotonic. QED.
