Does $f'(a)>0$ when $a<b$ imply that $f(a+b)>f(a) \impliedby a+b<c $

Question

Suppose I have some continiously differentiable function $f(a)$, such that $f'(a)>0$ if and only if $a<c$ ($c$ is some positive constant). Is it true that for all $b>0$, $$f(a+b)>f(a) \impliedby a+b<c?$$

My lecturer questioned this assumption. But it seems intuitively true: an epsilon increases in $a$ increase $f(a)$ when $a<c$. A series of epsilon increases should too, provided $a$ remains below $c$. Hence, a discrete change $b$ in the argument should increase $f(a)$ too. Is this true, and if so, why?

Answer 1

By the Mean Value Theorem, there is a point $x_0\in [a,a+b]$ such that $$f'(x_0)=f(a+b)-f(a)$$ But of course $x_0<c$ so $$f'(x_0)>0$$ which implies that $$f(a+b)-f(a)>0$$ and we are done.

Answer 2

If I understand the question[1], it is obvious.

Forget $a + b < c$ and using the same variable $a$ in two contexts. Let $d = a+b$ and you have $a < d < c$.

And you have the statement that $f'(x) > 0$ if and only if $x < c$[1]. Then $f$ is strictly increasing at all $x < c$.

So $f(a) < f(d) < f(c)$ when $a < d < c$.

That's it.

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[1] (I may not understand what you meant by "I have some continiously differentiable function f(a), such that f′(a)>0". It seems you are playing fast and loose with whether $a$ is constant or a variable. But either way $f'(a) > $ if and only if $a < c$ ought to mean that for all $x < c$, $f'(x) > 0$, oughtn't it?)