Answer
(a) $f(-1) = -1$
(b) $f(0) = 2$
(c) $f(2) = 6$
(d) $f(t^2+1) = 2t^2+4$
Domain: All real numbers
Range: $(-\infty, 1) ∪ [2,\infty)$
Work Step by Step
(a)
$f(-1) = 2(-1)+1,$ since $-1<0$
$f(-1)= -1$
(b)
$f(0) = 2(0)+2,$ since $0\geq0$
$f(0) = 2$
(c)
$f(2) = 2(2)+2,$ since $2\geq0$
$f(2) = 6$
(d)
$f(t^2+1) = 2(t^2+1)+2,$ since the $t^2+1\geq 0$
$f(t^2+1) = 2t^2+4$
Domain: All real numbers
Range: $(-\infty, 1) ∪ [2,\infty)$
