Condition For Local Extremum And A More Questions I am going to link a picture of the question (as well as a short solution/explanation which I did not understand).

Question
Let $f: [a, b] \to\mathbb R$ be a function such that for $c \in (a, b)$
$$f^i(c) = f^{ii}(c) = f^{iii}(c) = f^{iv}(c) = f^v(c) =
 0$$
where $f^n(c)$ represents the nth derivative of $f(c)$
Choose the correct options
  
  
*
  
*$f$ has a local extremum at $x = c$
  
*$f$ has neither a local extremum nor a local minimum at $x = c$
  
*$f$ is necessarily a constant function
  
*none of these —> (this is the correct option)
  
  
  Explanation
('Because' symbol here please. Couldn't find it on my keyboard) $$f^i(c) = f^{ii}(c) = f^{iii}(c) = f^{iv}(c) = f^v(c) = 0$$
Now, if $n$ is the least positive integer such that $f^n(c) \ne 0$,
  then it is not clear whether $n$ is even or odd. So, nothing can be
  said whether $f(x)$ has local extremum at $x = c$ or $n = t$.


After you are done with the picture, here are my questions:


*

*What exactly is the condition for $f(x)$ to have a local extremum? Till now, I used to solve for $f'(x) = 0$ and then check the sign of $f''(x)$ at that point. Do I need to check higher order derivates of $f(x)$ as well?

*If $f'(a) = f''(a) = 0$, then does it mean that there is neither a maximum nor a minimum at $x = a$?

*Can there be global/local maxima at boundary points of the domain? For eg: If the domain of $f(x)$ is $(0,50)$ and it attains maximum value at $x = 50,$ then does it count as a maximum?

*What is the condition for $f(x)$ to be a constant function? I think that all higher order derivatives of $f(x)$ should be zero. Is this correct?

Edit
Currently, I am banned from asking more questions because I ask "bad" questions, i.e., questions which got downvoted (a couple of such questions including this one) primarily because they were poorly formatted. So, I have spent a little over half an hour trying to improve this question (the other one has since been formatted by a more experienced user but still has a negative score) and I would appreciate a few upvotes so that I can ask new questions. ——> Update : I can now ask new questions. Thanks everyone !! 
 A: Based on the derivative test, if $f''(a)=0$, the test is inconclusive.
Referring to your example, consider:
$1) f(x)=x^6$ for $x\in [-1,1]$. Then:
$$f'(0)=f''(0)=f'''(0)=f^{(4)}(0)=f^{(5)}(0)=0, \\
f^{(6)}(0)=6!\ne 0; \\
f(0)=0 \text{ (min)}$$
$2) f(x)=x^7$ for $x\in [-1,1]$. Then:
$$f'(0)=f''(0)=f'''(0)=f^{(4)}(0)=f^{(5)}(0)=f^{(6)}(0)=0, \\
f^{(7)}(0)=7!\ne 0; \\
f(0)=0 \text{ (not max, not min)}$$
A: The "explanation" seems to assume that you know a theorem that says something about the least positive integer $n$ for which $f^{(n)}(c)$ is $0,$ and whose conclusion says something about whether there is a maximum or minimum or neither of the above. It might be better for you to include the theorem in your question.
Here is an application to the situation where $f'(c)=f''(c) = f'''(c)=0$ and $f''''(c)>0.$ If the function is reasonably well behaved, then $f''''>0$ in some neighborhood of $c,$ so  Then the third derivative is INCREASING in that neighborhood, so it must go from negative to positive. Since $f'''<0$ to the left of $c,$ $f''$ must be decreasing there, and likewise $f''$ is increasing to the right of $c.$ Thus $f''>0$ near $c,$ except exactly at $c.$ Therefore $f$ is concave upward near $c.$ Therefore there is a local minimum at $c.$
There is a theorem that says that if $f$ has first and second derivatives in some open interval containing $c$ and $f'(c)=0$ and $f''(c)>0,$ then $f$ has a local minimum at $c,$ and if $f'(c)=0$ and $f''(c)<0,$ then $f$ has a local maximum at $c.$ But that theorem says nothing about what happens when $f'(c)=0$ and $f''(c)=0.$ That is the situation treated by some theorems dealing with values at $c$ of higher derivatives of $f.$ There are some functions $f$ for which $f'(c) = 0 = f''(c)$ and there is a local minimum at $c.$ A very simple example is $f(x) = x^4$ and $c=0.$ There are others for which there is a maximum, and one example is $f(x) = -x^4$ and $c=0.$ And there are some for which there is no local extremum (i.e. maximum or minimum). $f(x) = x^5$ and $c=0$ is an example.
To say that you always have to look at the first and second derivatives is false. It can be done that way, but in some cases you can use only the first derivative and in some cases no derivative at all. You don't need any derivatives in order to know that $f(x) = x^2$ has a global minimum at $x=0,$ or that $f(x) = \sin x$ has a global maximum at $x = \pi/2.$ And sometimes only the first derivative is enough; for example, if $f$ is continuous and you show that $f'>0$ on the interval $(-\infty,c)$ and is negative on the interval $(c,+\infty),$ then you can conclude that $f$ has a global maximum at $c$ without knowing anything about the second derivative.
