Suppose f is differentiable on R and a≤f′(x)≤b for all x∈R where a,b>0. If f(0)=0, then

For x > 0.   Applying Lagrange’s theorem on [0, x]  we have c  ∈  (0,  x)such that f(x)x  =  f(x)−f(0)x−0  =  f'(c) But a≤f′(c)≤b so a≤f(x)x≤b⇒ax≤f(x)≤bx,x>0 Similarly for x<0_ applying Lagrange's theorem for [x,0⌉,  we have ax≤f(x)≤bx