Chapter 1.1, Problem 46E

To determine

To evaluate: $g\left(b^2+1\right)$, if $g\left(x\right)=\frac{3-x}{5+x}$ .

Answer to Problem 46E

  $\frac{2-b^2}{6+b^2}$

Explanation of Solution

Given information: The given value of $g\left(x\right)$ is as follows:

  $g\left(x\right)=\frac{3-x}{5+x}$ … (1)

Calculation:

To evaluate $g\left(b^2+1\right)$ , substitute the value of $x=b^2+1$ in equation (1) and simplify as shown below.

  $\begin{array}{c}g\left(x\right)=\frac{3-x}{5+x}\\ g\left(b^2+1\right)=\frac{3-\left(b^2+1\right)}{5+\left(b^2+1\right)}\\ =\frac{2-b^2}{6+b^2}\end{array}$

Hence, the value of $g\left(b^2+1\right)$ is $\frac{2-b^2}{6+b^2}$ .