Use the mean value theorem to show that |f(x)|<=(8x^2)/3 for 0<=x<=1/2.
My attempt was this:
f'(x)=-2x/(1-x^2), and so the is some c such that 0<=c<=t, and
|f(t)|=2ct/(1-c^2). But since 2c/(1-c^2) is increasing on 0<=c<=t, then
Is this right? Am I going in the right direction? I'm not sure where to go from here.