## Calculus, 10th Edition (Anton)

(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$}. \end{cases}$
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: $$g(t^{2} - 1)=\begin{cases} \pm t, & \text{t ^ {2} \geq 2}.\\ 3, & \text{t^{2} < 2}. \end{cases}$$