Suppose that $f(x)$ is a real continuous function defined in $0 \leq x \leq 1$ and bounded such that $a \leq f(x) \leq b$ where $a<b$
Suppose also that the derivative $f'(x)$ is also continuous in $0 \leq x \leq 1$ and bounded such that $c \leq f'(x) \leq d$ where $c<d$.
Is there any continuous transformation (linear or non linear) that I can apply to $\bf{{f(x)}}$ such that $-1 \leq f'(x) \leq 1$
Yes, you can do this.
Write $g(x) = rf(x) + sx$ for some to-be-determined constants $r,s$ with $r>0$. Then you know $$\begin{align} g'(x) &= r f'(x) + s \\ f'(x) &= \frac{g'(x)-s}{r} \\ c &\le \frac{g'(x)-s}{r} \le d \\ cr+s &\le g'(x) \le dr+s \end{align}$$
From there, you just need to solve a system of equations to find $r,s$ so that $cr+s = -1$ and $dr+s = 1$. Then you'll have $-1 \le g(x) \le 1$, so $g(x)$ will be the transformed version of $f$ you wanted.
