Question

Find the value of $g\left(-2\right),g\left(2\right),g\left(0\right),g\left(a\right),g\left(a^2-2\right),g\left(a+1\right)$for the function $g\left(t\right)=\frac{t+2}{t-2}$.

Short answer

The value of function are

$g\left(-2\right)=0,g\left(2\right)=\text{undefined},g\left(0\right)=1,g\left(a\right)=\frac{a+2}{a-2},g\left(a^2-2\right)=\frac{a^2}{a^2-4},g\left(a+1\right)=\frac{a+3}{a-1}$.

Step-by-step solution

Step 1. Evaluate g−2,g2,g0.

From the given function,

$g\left(t\right)=\frac{t+2}{t-2}\mathrm{....}\left(1\right)$

Evaluate the value of function at $t=-2,2,0$gives,

role="math" localid="1644503368684" $$\begin{gathered} g\left(-2\right)=\frac{\left(-2\right)+2}{\left(-2\right)-2}\mathrm{....}\left(\text{Substitute}t=-2\text{in}\left(1\right)\right) \\ =0 \\ g\left(2\right)=\frac{\left(2\right)+2}{\left(2\right)-2}\mathrm{....}\left(\text{Substitute}t=2\text{\hspace{0.17em}in}\left(1\right)\right) \\ =\frac{4}{0}\mathrm{....}\left(\text{Undefined}\right) \\ g\left(0\right)=\frac{\left(0\right)+2}{\left(0\right)-2}\mathrm{....}\left(\text{Substitute}t=0\text{in}\left(1\right)\right) \\ =-1 \end{gathered}$$

Step 2. Evaluate ga,ga2−2,ga+1 for the given function.

Similarly, evaluate the value of function (1) at$t=a,a^2-2,a+1$ gives,

role="math" localid="1644503713306" $$\begin{gathered} g\left(a\right)=\frac{\left(a\right)+2}{\left(a\right)-2}\mathrm{....}\left(\text{Substitute}t=a\text{in}\left(1\right)\right) \\ =\frac{a+2}{a-2} \\ g\left(a^2-2\right)=\frac{\left(a^2-2\right)+2}{\left(a^2-2\right)-2}\mathrm{....}\left(\text{Substitute}t=a^2-2\text{\hspace{0.17em}in}\left(1\right)\right) \\ =\frac{a^2}{a^2-4} \\ g\left(a+1\right)=\frac{\left(a+1\right)+2}{\left(a+1\right)-2}\mathrm{....}\left(\text{Substitute}t=a+1\text{in}\left(1\right)\right) \\ =\frac{a+3}{a-1} \end{gathered}$$

Step 3. Description of steps.

From the obtained values,

$g\left(-2\right)=0,g\left(2\right)=\text{undefined},g\left(0\right)=1,g\left(a\right)=\frac{a+2}{a-2},g\left(a^2-2\right)=\frac{a^2}{a^2-4},g\left(a+1\right)=\frac{a+3}{a-1}$.