Let $f$ be a real-valued function continuous on an interval $I$ and differentiable on every interior point of $I.$ Then if $f'(x)\ge 0$ everywhere inside $I,$ $f$ does not decrease over $I.$
There's this proof using MVT: Let $x<y$ be arbitrary numbers in $I,$ then there is a $c\in (x,y)$ such that $$f'(c)=\frac{f(y)-f(x)}{y-x}.$$ But since $f'(c)\ge 0$ and $y-x>0,$ this gives $f(y)-f(x)=f'(c)(y-x)\ge 0,$ so that $f(y)\ge f(x)$ on $I.$
I tried proving the contrapositive, and ended up with something quite unsatisfactory. Here's what I did. I assumed that $f$ decreases on $I,$ then applied MVT too: Thus, with notation as above, we have $$f'(c)=\frac{f(y)-f(x)}{y-x}$$ for arbitrary $c\in (x,y),$ with $x$ and $y$ also being arbitrary. However, since $f$ decreases, it follows that $f'(c)<0,$ which is the conclusion we should have, too.
However, I noticed that this fails for some example. I just applied this to $-x^3,$ which decreases everywhere, but whose derivative $-3x^2$ fails to be negative at $x=0,$ as expected. What is happening here?
As a sanity check, this does not happen for $x^3$ using the first version of the theorem. So, what is wrong with the contrapositive, or its proof?
Thanks.
Note that if "$f$ decreases on $I$", then "$f'(x) \ge 0$ everywhere inside of $I$" no longer fully applies. In particular, this hypothesis means that $f'(x) \lt 0$ at least somewhere in $I$, but not necessarily everywhere. With your example of $f(x) = -x^3$, even though it's always decreasing over any range, you've seen that the derivative may be $0$, which is due to it being an inflection point there, even though the derivative is negative everywhere else.