Chapter 0.5, Problem 40E

To determine

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

Answer to Problem 40E

The domain and range of the function $y=-3\log \left(x+2\right)+1$ are $\left(-2,\infty \right)$ and $\left(-\infty ,\infty \right)$ , respectively.

Explanation of Solution

Given data:

The function is $y=-3\log \left(x+2\right)+1$ .

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 $f\left(x\right)=-3\log \left(x+2\right)+1$ .

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

  

Clearly, the function $f\left(x\right)$ is defined only when $\left(x+2\right)\ge 0$ or $x\ge -2$ . Therefore, the domain of the function $f\left(x\right)=-3\log \left(x+2\right)+1$ is interval $\left(-2,\infty \right)$ .

Further,

  $\begin{array}{c}y=-3\log \left(x+2\right)+1\\ \log \left(x+2\right)=\frac{1-y}{3}\\ e^{\left(1-y\right)/3}=x+2\\ =\frac{e^{1/3}}{e^{y/3}}=x+2\end{array}$

If $\left(x+2\right)\ge 0$ , then $\frac{e^{1/3}}{e^{y/3}}\ge 0$ . But $\frac{e^{1/3}}{e^{y/3}}\ge 0$ 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 $y=-3\log \left(x+2\right)+1$ .

Hence, the range of the function $y=-3\log \left(x+2\right)+1$ is $\left(-\infty ,\infty \right)$ .