Answer
There are no critical points.
Work Step by Step
$f'(t)=4-\frac{1}{2\sqrt {t^{2}+1}}(2t)$
$=\frac{8\sqrt {t^{2}+1}-2t}{2\sqrt {t^{2}+1}}$
As the denominator is always not equal to 0 and $t^{2}+1$ inside the square root is always positive, $f'(t)$ exists for all $t$.
$f'(t)=0\implies8\sqrt {t^{2}+1}-2t=0$
$\implies64(t^{2}+1)-4t^{2}=0$
Or, $60t^{2}=-64$
There are no solutions for this equation.
The critical points are those $t$ for which$f'(t)=0$ or $f'(t)$ doesn't exist. As there are no such points, this function has no critical points.
