Chapter 0.5, Problem 39E

To determine

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

Answer to Problem 39E

The domain and range of the function $y=2\ln \left(3-x\right)-4$ are $\left(-\infty ,3\right)$ and $\left(-\infty ,\infty \right)$ , respectively.

Explanation of Solution

Given data: The given function is $y=2\ln \left(3-x\right)-4$ .

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 $f\left(x\right)=2\ln \left(3-x\right)-4$ .

The graph of the function $f\left(x\right)=2\ln \left(3-x\right)-4$ is shown in Fig. 1.

  

Clearly, the function $f\left(x\right)$ is defined only when $\left(3-x\right)\ge 0$ or $x\le 3$ . Therefore, the domain of the function $f\left(x\right)=2\ln \left(3-x\right)-4$ is interval $\left(-\infty ,3\right)$ .

Further,

  $\begin{array}{c}y=2\ln \left(3-x\right)-4\\ \ln \left(3-x\right)=\left(y+4\right)/2\\ \Rightarrow e^{\left(y+4\right)/2}=\left(3-x\right)\end{array}$

If $\left(3-x\right)\ge 0$ , then $e^{\left(y+4\right)/2}\ge 0$ but $e^x$ is always positive whatever be the value of x . Therefore, $e^{\left(y+4\right)/2}\ge 0$ for every value $y\in \left(-\infty ,\infty \right)$ . Hence, the range of the function $y=2\ln \left(3-x\right)-4$ is $\left(-\infty ,\infty \right)$ , that also clear from the Fig. 1.