Question

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

Short answer

The value of function are

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

Step-by-step solution

Step 1. Evaluate g2,g−1,g12 for the given function.

From the given function,

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

Evaluate the value of function at $x=2,-1,\frac{1}{2}$gives,

role="math" localid="1644501947894" $$\begin{gathered} g\left(2\right)=\frac{1-\left(2\right)}{1+\left(2\right)}\mathrm{....}\left(\text{Substitute}x=2\text{in}\left(1\right)\right) \\ =-\frac{1}{3} \\ g\left(-1\right)=\frac{1-\left(-1\right)}{1+\left(-1\right)}\mathrm{....}\left(\text{Substitutex}=-1\text{\hspace{0.17em}in}\left(1\right)\right) \\ =\frac{2}{0}\mathrm{....}\left(\text{Undefined}\right) \\ g\left(\frac{1}{2}\right)=\frac{1-\left(\frac{1}{2}\right)}{1+\left(\frac{1}{2}\right)}\mathrm{....}\left(\text{Substitutex}=\frac{1}{2}\text{in}\left(1\right)\right) \\ =\frac{1}{3} \end{gathered}$$

Step 2. Evaluate ga,ga−1,gx2−1 for the given function.

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

role="math" localid="1644502213949" $$\begin{gathered} g\left(a\right)=\frac{1-\left(a\right)}{1+\left(a\right)}\mathrm{....}\left(\text{Substitute}x=a\text{in}\left(1\right)\right) \\ =\frac{1-a}{1+a} \\ g\left(a-1\right)=\frac{1-\left(a-1\right)}{1+\left(a-1\right)}\mathrm{....}\left(\text{Substitutex}=a-1\text{\hspace{0.17em}in}\left(1\right)\right) \\ =-1 \\ g\left(x^2-1\right)=\frac{1-\left(x^2-1\right)}{1+\left(x^2-1\right)}\mathrm{....}\left(\text{Substitutex}=x^2\text{-1in}\left(1\right)\right) \\ =\frac{2-x^2}{x^2} \end{gathered}$$

Step 3. Description of steps.

From the obtained values,

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