Theorem:
If $f(x)$ , a function of one variable , is continuous on an interval I and has only one critical point in I (where a critical point is a point c at which $f'(c)$ does not exist or $f'(c)=0$),and that critical point is a local minimum in I , then it is the global minimum in I.
(Same applies for maximum)
This theorem does not hold in higher dimensions
eg.$f(x,y)=e^{3x}+y^3-3ye^x$(You can show using second derivative test that it fails the theorem)
But it does hold in one dimension .
I want to know how to prove this theorem in one dimension(Whatever the proof is fails in higher dimensions)
To prove it i think you have to use some property of functions from analysis
Edit: if f is differentiable on (c,x] and the derivative is zero nowhere and has the same signed derivative on (c,x) then can you make any conclusion about the sign derivative of x or not?
Edit: The theroem that is critcial for the proof of this theorem is that If derivative has different signs at $a, b\in I$ then by applying Darboux on $[a, b] $ the derivative vanishes somewhere on $(a, b) $. So to apply to our case if the sign is different with any two points within (c,x) then the derivative must vanish in (c,x) which contracdicts the hypothesis hence the sign must be same for all (c,x]
Best Answer
By your assumption every point of $I$ except $c$ is not a critical point. By definition of critical point in your question, it means that if $x\in I, x\neq c$ then $f'(x) \neq 0$.
Let us deal with the case when $c$ is an interior point of $I$. It is also given that $c$ is a local minimum. Consider any $x\in I$ and $x>c$. The derivative $f'$ does not vanish in interval $(c, x) $ and hence by Darboux theorem maintains a constant sign. Now if this derivative $f'$ were negative in $(c, x) $ then we would have $f(y) <f(c) $ for all $y\in(x, c) $ and this contradicts that $c$ is a local minimum. It follows that $f'$ is positive in $(c, x) $ and hence $f(x) >f(c) $.
Similarly for $x\in I$ with $x<c$ we can prove that $f(x) > f(c) $ so that $c$ is indeed a global minimum on $I$.
The proof is simpler and similar if $c$ is an end point of $I$.