Given $f(x)=3x+2$ for $x<0$ and $f(x)= 3x +3$ for $x\ge0$, what is $f(t^2 + 1)$? I need to find $f(t^2 + 1)$ given $$f(x) = \begin{cases} 3x+2 & x \lt 0\\ 3x+3 & x\ge 0\end{cases}$$
The textbook says the answer is $3t^2 + 6$, but how do you know which piecewise to plug $t^2 + 1$ into?

$$t^2 + 1 \geq 1 \quad \forall t \in \mathbb R$$
(The square of any number is greater than or equal to $0$. So $t^2 + 1\geq 1$.)
So we use $f(x) = 3x + 3$. Then we have $$f(t^2 + 1) = 3(t^2+1) + 3 = 3t^2 + 6$$
