Chapter 0.5, Problem 41E

To determine

To Find: The domain and range of a function using graphing utility.

Answer to Problem 41E

The domain and range of the function $y={\log }_2\left(x+1\right)$ are $\left(-1,\infty \right)$ and $\left(-\infty ,\infty \right)$ , respectively.

Explanation of Solution

Given data:

The function is $y={\log }_2\left(x+1\right)$ .

Method/Formula used:

The domain of a function $y=f\left(x\right)$ is the set of values of x for which $f\left(x\right)$ is defined and the set of values of y for all the members of domain is the range of function $f\left(x\right)$ .

Calculation:

The given function is $y={\log }_2\left(x+1\right)$ .

First convert $y={\log }_2\left(x+1\right)$ at the base of 10 or ‘ e’ as follows:

  $\begin{array}{c}y={\log }_2\left(x+1\right)\\ \Rightarrow 2^y=x+1\end{array}$

Taking log of both sides at base ‘ e’

  $\left.\begin{array}{c}y\ln 2=\ln \left(x+1\right)\\ y=\frac{\ln \left(x+1\right)}{2}\end{array}\right\}$ ... (1)

The graph of the function $y=\frac{\ln \left(x+1\right)}{\ln 2}$ is shown in Fig. 1.

  

Clearly, the function $f\left(x\right)$ is defined only when $\left(x+1\right)\ge 0$ or $x\ge -1$ . Therefore, the domain of the function $y={\log }_2\left(x+1\right)$ is interval $\left(-1,\infty \right)$ .

Further,

  $\begin{array}{c}y={\log }_2\left(x+1\right)\\ 2^y=x+1\end{array}$

If $\left(x+1\right)\ge 0$ , then $2^y\ge 0$ . But $2^y\ge 0$ whatever be the value of y in the interval $\left(-\infty ,\infty \right)$ . Therefore, $y\in \left(-\infty ,\infty \right)$ that is also clear from the graph of function $y={\log }_2\left(x+1\right)$ .

Hence, the range of the function $y={\log }_2\left(x+1\right)$ is $\left(-\infty ,\infty \right)$ .