Chapter 5.3, Problem 52AYU

To determine

The graph of function $f(x)=1-2^{-x/3}$ using transformation. Also, domain, range, horizontal asymptote, $y$ - intercept of function.

Answer to Problem 52AYU

Solution:

The graph of $f(x)=1-2^{-x/3}$ is

  

Domain of the function $f(x)=1-2^{-x/3}$ is $\left(-\infty ,\infty \right)$

Range of function $f(x)=1-2^{-x/3}$ is $\left(-\infty ,1\right)$

Horizontal asymptote of $f(x)=1-2^{-x/3}$ is $y=1$

  $y$ - intercept of $f(x)=1-2^{-x/3}$ is $0$

Explanation of Solution

Given information:

The function $f(x)=1-2^{-x/3}$

To graph a function, begin with a graph of $f(x)=2^x$

Find some points satisfying $f(x)=2^x$

  $\begin{array}{cccccc}x& -1& 0& 1& 2& 3\\ f\left(x\right)& \begin{array}{l}f\left(-1\right)=2^{-1}\\ =\frac{1}{2}\end{array}& \begin{array}{l}f\left(0\right)=2^0\\ =1\end{array}& \begin{array}{l}f\left(1\right)=2^1\\ =2\end{array}& \begin{array}{l}f\left(2\right)=2^2\\ =4\end{array}& \begin{array}{l}f\left(3\right)=2^3\\ =8\end{array}\\ \left(x,f\left(x\right)\right)& \left(-1,\frac{1}{2}\right)& \left(0,1\right)& \left(1,2\right)& \left(2,4\right)& \left(3,8\right)\end{array}$

Plot these points on the graph.

Therefore, graph is

  

Using the above graph, take reflection about $y$ axis to get graph of $f(x)=2^{-x}={\left(\frac{1}{2}\right)}^x$

The graph is

  

To graph a function $f(x)=-2^{-x/3}$ , stretch the graph horizontally of $f(x)=2^{-x}={\left(\frac{1}{2}\right)}^x$ by $1/3$ units and reflect it about $x$ axis

The graph of $f(x)=2^{-x/3}$ is

  

To graph the function $f(x)=1-2^{-x/3}$ , shift the graph of $f(x)=-2^{-x/3}$ upward by $1$ unit.

The graph of $f(x)=1-2^{-x/3}$ is

  

From the graph, domain of the function $f(x)=1-2^{-x/3}$ is $\left(-\infty ,\infty \right)$

Range of the function is the set of values of the dependent variable for which a function is defined.

Therefore, range of $f(x)=1-2^{-x/3}$ is $\left(-\infty ,1\right)$

As $x\to \infty$ , $f(x)$ approaches to $1$

Therefore, $f(x)=1-2^{-x/3}$ has horizontal asymptote $y=1$

To find $y$ - intercept, substitute $x=0$ in $f(x)=1-2^{-x/3}$

  $\Rightarrow f(0)=1-2^{-0/3}=1-1=0$

Therefore, $y$ - intercept is $0$