Theorem 5.8 Baby Rudin. Some questions This is with reference to the page number 107 of Baby Rudin. 

Theorem 5.8 Let $f$ be defined on $[a,b]$.If a point $x\in(a,b)$ is a
  local maximum of function $f$, and if $f'(x)$ exist, then $f'(x)=0$.

I have some questions. Thanks in advance for reading and helping out.


*

*What about the converse of this theorem? Suppose $f'(x)=0$ at some point, can we always conclude that it is local maximum? 

*The theorem add a condition that $f'(x)$ should exist. This raised a question, Can we have a function function with local maximum/minimum at some point say $p$ but derivative undefined at the point.
Edits: 
We can take $f(x)=x^2$ on $[-1,1]$. $f'(0)=0$ but $0$  is not local maximum. I think this works for (1).
 A: Answer to question 1
The derivative of $f(x)=x^3$ vanishes at $0$ but $f$ doesn’t have a minimum nor at maximum at $0$.
Answer to question 2
Take $g(x)=\vert x \vert$, again look at zero.
A: If $f'(x)=0$, then $x$ is said to be a stationary point. It can be something like $x=0$ in $f(x)=x^3$ too, for example.
For (2), we can even have a function with a global maximum at $p$ but without a derivative in this point. See, for example, $f(x)=-|x|$.
A: 
  
*
  
*What about the converse of this theorem? Suppose $f'(x)=0$ at some point, can we always conclude that it is local maximum? 
  

No. It could be a minimum, or it could be neither. Such examples:


*

*$f(x) = x^2$. Here, $f'(x) = 2x$ thus implying $f'(x) = 0$ for $x = 0$. It is immediately clear on looking at the graph that this is a local minimum for some intervals of the function.

*$f(x) = x^3$. Here, $f'(x) = 3x^2$. Thus, $f'(x) = 0$ only if $x=0$. On looking at the graph, it is clearly not a local extremum in either respect, up to choice of the interval.
(I say "up to choice of the interval" because $x=0$ yields a local minimum on the choice of interval $[0,1]$ for both functions, and on $[-1,0]$ for example the latter has a maximum instead. If we generally state intervals $[a,b]$ with $a<0<b$, though, they respectively have a minimum and neither respectively at $x=0$.)
This touches on the nature of "critical points" you might have learned in your introductory calculus courses: points where $f'(x) = 0$ or is undefined are candidates for local extrema.



  
*The theorem add a condition that $f'(x)$ should exist. This raised a question, Can we have a function with local maximum/minimum at some point say $p$ but derivative undefined at the point?
  

Yes. Consider $f(x) = |x|$ on the interval $[a,b]$ where $a<0<b$. The function clearly has a minimum at $x=0$, but the derivative is undefined at that point.
A: Let me try:
1) $f'(x)=0; $
Example:$ f(x)=x^3$, at $x= 0$,  inflection point, not an extremum
2) $f(x)= |x|$ , has a local minimum at $x=0.$
$f(x)= -|x|$ has a local maximum at $x=0$
(f'(0) does not exist)(Why?)
