Let f(x)=log(1-x^2)

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

-2c/(1-c^2)=log(1-t^2)/t, so,

|f(t)|=2ct/(1-c^2). But since 2c/(1-c^2) is increasing on 0<=c<=t, then

|f(t)|<=2t^2/(1-t^2).

Is this right? Am I going in the right direction? I'm not sure where to go from here.