Given that an extreme value can only occur at a critical point and in the following case $$f(x) = \frac{1}{x} \quad [1,4]$$ we definitely have two absolute extreme values (maximum and minimum), are $x = 1$ and $x = 4$ critical points?
If so, what is the reasoning for that: the fact that they are in the domain of $f$ and they are not differentiable at those points?
Is my reasoning correct?
On
I want to point out that it is not sure that you have an extreme point in the critical points. One example is $f(x)=x^3$ which has a $x=0$ as critical point but obviously it's not an extreme.
When we are trying to find a critical point in a certain domain we set $f'(x)=0$. Then (for the fact that I mentioned that not every critical point is extreme) we plug those critical points and the endpoints of the given domain(in case this domain has endpoints)in the function and see which of them is the smaller value(minimum) and which the bigger(maximum). You realise that altough the endpoints may not be critical points, they can behave as extreme points. That happens to the function you mention.
A critical point is a point at which the derivative vanishes. So definitely, $1$ and $4$ are not critical points.
Now those points are at the boundary of the domain of $f$ and are extremas.
However, consider a point $x$ which is a minimum or a maximum of a differentiable function $f$ and which belongs to the interior of the the domain of $f$. Then $f^\prime(x)=0$.
In summary
An extrema belonging to the interior of the domain of a differentiable map is a critical point.
An extrema may not be a critical point, if it belongs to the frontier of the domain. Example: the function of the question.
And obviously the derivative of a function can vanish at a point belonging to the frontier. In that case the point is a critical point.
Lastly a critical point may not be an extrema. Example $f: x \mapsto x^3$ at $x=0$.