$\text{ Suppose }\mathrm{f}\text{ is differentiable on }\mathrm{R}\text{ and }a\le f^{\mathrm{\prime }}\left(x\right)\le b\text{ for all }x\in R\text{ where }$$a,b>0.\text{ If }f\left(0\right)=0,\text{ then }$

For x > 0.   Applying Lagrange’s theorem on [0, x]  we have $c\in (0,x)$
such that $\frac{f\left(x\right)}{x}=\frac{f\left(x\right)-f\left(0\right)}{x-0}=f\text{'}\left(c\right)$

$\begin{array}{l}\text{ But }a\le f^{\mathrm{\prime }}\left(c\right)\le b\text{ so }a\le \frac{f\left(x\right)}{x}\le b\Rightarrow ax\le f\left(x\right)\le bx,x>0\\ \text{ Similarly for }\mathrm{x}<\underset{\_}{0}\text{ applying Lagrange's theorem for }[\mathrm{x},0\rceil \text{, }\\ \text{ we have }ax\le f\left(x\right)\le bx\end{array}$