If ${\text{⨍ }}^{\text{'}}\text{(x)=}\frac{1}{1+{\mathrm{x}}^2}$ for all x and $\text{⨍ (0)=0,}$ then :

$$\begin{gathered} {\mathrm{f}}^1\left(\mathrm{x}\right)=\frac{\mathrm{f}\left(2\right)-\mathrm{f}\left(0\right)}{2-0}=\frac{\mathrm{f}\left(2\right)}{2} \\ \mathrm{f}\left(2\right)=\frac{2}{1+{\mathrm{x}}^2} \\ 0<\mathrm{x}<2 \\ \Rightarrow 0<{\mathrm{x}}^2<4 \\ 1<1+{\mathrm{x}}^2<5 \\ 0.2<\frac{1}{1+{\mathrm{x}}^2}<1 \\ 0.4<\frac{2}{1+{\mathrm{x}}^2}<2 \\ 0.4<\mathrm{f}\left(2\right)<2 \end{gathered}$$