Chapter 5, Problem 1RE

To determine

(a)

To determine:

Find the function values (if they exist),$k(2),k(0),k(1),k(-1),k\left(\frac{1}{2}\right)$

Answer to Problem 1RE

Solution:

$\begin{array}{l}k\left(2\right)=\frac{2}{3}\\ k\left(0\right)=0\\ k\left(1\right)=\infty \\ k\left(-1\right)=\infty \\ k\left(\frac{1}{2}\right)=\frac{-2}{3}\end{array}$

Explanation of Solution

Given info.

$k\left(y\right)=\frac{y}{y^2-1}$

Calculation:

(a) Substituting given value=$2$ in function

$\begin{array}{c}k\left(2\right)=\frac{2}{2^2-1}\\ =\frac{2}{4-1}\\ =\frac{2}{3}\end{array}$

Therefore, the value of $k\left(2\right)$ is $\frac{2}{3}$

Substituting $0$ in function $k\left(y\right)=\frac{y}{y^2-1}$

$\begin{array}{c}k\left(0\right)\text{ = }\frac{0}{0^2-1}\\ =\text{ 0}\end{array}$

Substituting $1$ in function $k\left(y\right)=\frac{y}{y^2-1}$

$\begin{array}{c}k\left(1\right)=\frac{1}{1^2-1}\\ =\frac{1}{0}\\ =\infty \end{array}$

Substituting $-1$ in function $k\left(y\right)=\frac{y}{y^2-1}$

$\begin{array}{c}k\left(-1\right)=\frac{-1}{{\left(-1\right)}^2-1}\\ =\frac{-1}{1-1}\\ =\frac{-1}{0}\\ =\infty \end{array}$

Substituting $\frac{1}{2}$ in function $k\left(y\right)=\frac{y}{y^2-1}$

$\begin{array}{c}k\left(\frac{1}{2}\right)=\frac{\frac{1}{2}}{{\left(\frac{1}{2}\right)}^2-1}\\ =\frac{\frac{1}{2}}{\frac{1}{4}-1}\\ =\frac{\frac{1}{2}}{\frac{1-4}{4}}\end{array}$

$=\frac{\frac{1}{2}}{\frac{-3}{4}}\text{ Convert the fraction into Proper numerator and denominator}$

$\begin{array}{c}\text{= }\frac{\text{1}}{\text{2}}\text{ × }\frac{\text{4}}{-\text{3}}\end{array}$

$\begin{array}{c}\text{On simplification, you get}\end{array}$

$\begin{array}{c}\text{= }\frac{-\text{2}}{\text{3}}\end{array}$

Therefore, the value of $k\left(\frac{1}{2}\right)is\frac{-2}{3}$

Conclusion:

$\begin{array}{l}k\left(2\right)=\frac{2}{3}\\ k\left(0\right)=0\\ k\left(1\right)=\infty \\ k\left(-1\right)=\infty \\ k\left(\frac{1}{2}\right)=\frac{-2}{3}\end{array}$

b)

Identify the domain for k write the answer in interval solution