# If F'(x) is bounded so is F(x)

Hello everybody,
A few years ago i tried to join a mathematics department and in the relevant exams
i came up against the following problem. I apologise beforehand if the statement of the problem is a little bit ambiguous because i do not remember it exactly. However, I am sure you will get the point.

## Homework Statement

if the first derivative of the real function F(x) is continuous and bounded over the interval [a,b] (or (a,b) ?) , prove that F(x) also is bounded on the interval (a,b) (or [a,b] ?) and the vice versa.

## Homework Equations

So we can see that m =< F'(x) =< M.
How can we get from this into the boundness of the F(x) without falling into pitfalls ?
What about the vice versa ?
Should we use the defintion or something else ?

## The Attempt at a Solution

I will not attempt to publish the solution I proposed because many of you may laugh. Related Calculus and Beyond Homework Help News on Phys.org
For starters, for the "vice-versa" thing to have a fighting chance, it would have to be on a closed interval. For example, x^1/2 is bounded on (0,1] but its derivative is not.

Indeed the "vice-versa" does not hold for every function F(x) as it has also been discussed https://www.physicsforums.com/showthread.php?t=515616&highlight=bounded+function+derivative".

How can we argue that F(x) is bounded ? By intuition the statement is obvious ...

Is integrating over (a,t) where t a variable among a and b the inequality m =< F'(x) <= M a good argument ? We will get rid off the derivative by doing so ... I have an engineering background so make allowances ...

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