Answer
$(\frac{11}{8},\frac{3}{4})$
Work Step by Step
$f(t)=t^2+t+1$
Find the derivative:
$f'(t)=2t+1$
$f'(t)=0$ when $t=\frac{1}{2}$
Also $f''(t)=2$
Hence, by the second derivative test, $f(t)$ is a minimum when $t=\frac{1}{2}$. The point correspondsing to $t=-\frac{1}{2}$ is $(\frac{11}{8},\frac{3}{4})$
