Answer
(a) $g(3) = 2$
$g(-1) = 0$
$g(\pi) = \frac{\pi + 1}{\pi - 1}$
$g(t^{2} - 1) = \frac{t^{2}}{t^{2} - 2}$
(b) $g(3) = \pm2$
$g(-1) = 3$
$g(\pi) = \sqrt{\pi + 1}$
$
g(t^{2} - 1)=\begin{cases}
\pm t, & \text{$t ^ {2} \geq 2$}.\\
3, & \text{$t^{2} < 2$}.
\end{cases}
$
Work Step by Step
To get the answers substitute the value for x into the given function:
(a) $g(3) = \frac{3 + 1}{3 - 1} = \frac{4}{2} = 2$
$g(-1) = \frac{(-1) + 1}{(-1) - 1} = \frac{0}{-2} = 0$
$g(\pi) = \frac{\pi + 1}{\pi - 1}$
$g(t^{2} - 1) = \frac{(t^2 - 1) + 1}{(t^2 - 1) - 1} = \frac{t^2}{t^2 - 2}$
For (b), it is necessary to work out which piece of the piecewise function to use, depending on the condition for each piece:
(b)For $x = 3$, as $3 \geq 1$, the first part of the function is used and $g(3) = \sqrt {x + 1} = \sqrt {3 + 1} = \sqrt4 = \pm 2$
For $x = -1$, as $-1 < 1$, the second part of the function is used and $g(-1) = 3$
For $x = \pi$, as $\pi \geq 1$, the first part of the function is used and $g(\pi) = \sqrt{\pi + 1}$
For $x = t^2 -1$, we do not know which of the conditions in the piecewise function so we have to consider both. If $t^2 - 1 \geq 1$, tis also means that $t^2 \geq 2$.If that condition is met, then $g(x) = \sqrt{x+1}$ and $g(t^2 - 1) = \sqrt {(t^2 - 1) + 1} = \sqrt {t^2} = \pm t$
However if $t^2 - 1 < 1$, this means that $t^2 < 2$. If that condition is met, then $g(x) = 3$.
Putting these parts of the function together we get: $$
g(t^{2} - 1)=\begin{cases}
\pm t, & \text{$t ^ {2} \geq 2$}.\\
3, & \text{$t^{2} < 2$}.
\end{cases}
$$
