Solution: To find the maximum value of $ f(t) = t^2 - \fract^44 $, take the derivative $ f'(t) = 2t - t^3 $. Set $ f'(t) = 0 $: $ 2t - t^3 = 0 \Rightarrow t(2 - t^2) = 0 $. Critical points at $ t = 0 $ and $ t = \pm\sqrt2 $. Evaluate $ f(\sqrt2) = (\sqrt2)^2 - \frac(\sqrt2)^44 = 2 - \frac44 = 1 $. The maximum value is $ \boxed1 $. - Appfinity Technologies