Applications of $f(x_0)=f'(x_0)$ If a function $f(x)$ has a derivative $f'(x)$ then where $f'(x_0) = 0$ there is an extreme point at $x=x_0$.
And where $f''(x_0)=0$ there is an inflection point at $x=x_0$.
I am asking are there any significance or applications where $f(x_0)=f'(x_0)$ at some point $x=x_0$?
I am referring to a case like $f(x)=x^2$ and $f'(x)=2x$, they intersect at $(0,0)$ and $(2,4)$ what is special or can be derived from these points?
 A: If I'm interpreting this correctly, I believe you are asking if there is anything special about this happening locally, i.e at a certain point $x_0$, does the fact that $f(x_0)=f'(x_0)$ mean anything? I would say no; note that by just translating the function appropriately, you can make this happen at any point. To see this, for any differentiable $f$, and any point $x_0$ define $g_{x_0}(x)=f(x)+(f'(x_0)-f(x_0))$, and this function will have that property at the point $x_0$. 
A: Say this happens at the point $(a,b)$. Then the slope of the tangent line that point is also $b$, so the tangent line at that point is
$$
y-b=b(x-a)
$$
which can be rewritten as
$$
y=b(x-(a-1))
$$
That is, the tangent line at $(a,b)$ is the line of slope $b$ which intersects the $x$-axis at the point $(a-1, 0)$.
So, if you linearly approximate $f$ based on its behavior at $(a,b)$, your approximation will vanish one $x$-unit before $a$. In some cases, this might mean your chosen units for $x$ and $f(x)$ are in some way compatible near $(a,b)$. However, it can't mean anything more physical than that, because if you change units then this condition will no longer be true!
A similar question you could ask that might be a little more meaningful (because it does continue to be true after a change of units) is:

What is the significance of a point $x_0$ where $f(x_0)=x_0f'(x_0)$?

By the same kind of computation I did above, you can check that this means the tangent line to $f$ at $x_0$ passes through the origin.
A: If $f(x_0)=f'(x_0)$, this tells us that if we define $g(x)=f(x)\exp(-x)$, that $g$ has an extreme point at $x_0$ (excepting inflection points etc.). This is because:
$$\frac{dg}{dx}=\frac{d}{dx}f(x)\exp(-x)=f'(x)\exp(-x)-f(x)\exp(-x)$$
$$\implies\frac{dg}{dx}\vert_{x=x_0}=f'(x_0)\exp(-x_0)-f(x_0)\exp(-x_0)=0$$
i.e. $g'(x_0)=0$.
A: If $f(x)=f'(x)$, then $f(x) = C \cdot e^x$.
That's the important application/consequence here.    
A: Well, $f(x) = f'(x)$, for every $x \implies f(x) = Ce^x$, C is a constant that can be determined if you have a inicial condition. 
There are many applications, such as radioactive decay or growth of some population!
