I am having a hard time with exercises of the form : $f'$ verify some properties then prove that $f$ is such that : ...
The main problem I have is that in order to link $f$ and it's derivative I only know that : $f(x) = f'(a)(x-a) + o(x)$ in a neighboorhhod of $a$. Yet this formula is not precise at all and only usefull to compute limit. The $o(x)$ term makes it impossible to manipulate/do $\epsilon$ proofs. So is there a way to have a precise formula that link $f$ and it's derivative so that it's more suitable for $\epsilon$ proof ?
For example let's consider the following :
Let $f \in C^1(\mathbb{R}, \mathbb{R})$. Moreover we know that on $[a,b]$ $\inf \mid f' \mid = K$ then prove that $\mid f\mid \geq Kx $ on $[a,b]$
This is intuitively obvious. Yet using the formula $f(x) = f'(a)(x-a) +o(x)$ doesn't help at all since it's only an asymptotic.
Thus is there a way/formula that relates precisely the behaviour of $f$ and the behaviour of $f'$ so that $\epsilon$ rpoofs are possible ?
Thank you !
$$|f'(x)|=K+\psi(x)~~\forall~x\in [a,b]$$
where $\psi(x)\geq 0$
If $f'(x)\lt 0$, then $$\begin{align}-f'(x)=K+\psi(x)&\implies -f(x)=Kx+\int\psi(x)~\mathrm dx\geq Kx+0=Kx\\&\implies |-f(x)|=|f(x)|\geq |K||x|=K|x|\geq Kx\end{align}$$
If $f'(x)\geq 0$, then $$\begin{align}f'(x)=K+\psi(x)&\implies f(x)=Kx+\int\psi(x)~\mathrm dx\geq Kx+0=Kx\\&\implies |f(x)|\geq |K||x|=K|x|\geq Kx\end{align}$$
So, in any case we have $|f(x)|\geq Kx~~\forall~x\in [a,b]$