Chapter 3.1, Problem 33E

Solution

To decide: if the given function is an exponential growth function or an exponential decay function and describe, its end behavior using limits.

Exponential decay function

  $\underset{x\to \infty }{\lim }f\left(x\right)=0$

  $\underset{x\to -\infty }{\lim }f\left(x\right)=\infty$

Given information:

The functions:

  $f\left(x\right)={0.5}^x$

Definition Used:

Exponential Growth and Decay:

For an exponential function $f\left(x\right)=a\cdot b^x$ and any real number x ,

  $f\left(x+1\right)=b\cdot f\left(x\right)$

If $a>0$ and $b>1$, the function f is increasing and is an exponential growth function. The base b is its growth factor.

If $a>0$ and $0<b<1$, the function f is decreasing and is an exponential decay function. The base b is its decay factor.

Explanation:

The given function is:

  $f\left(x\right)={0.5}^x$

Thus, by using the above definition for this function base is $b=0.5$ which satisfies $0<b<1$, hence this function is an exponential decay function.

Now, the end behavior of this function is:

  $\begin{array}{c}\underset{x\to \infty }{\lim }f\left(x\right)=\underset{x\to \infty }{\lim }{\left(0.5\right)}^x\\ =\underset{x\to \infty }{\lim }{\left(\frac{1}{2}\right)}^x\\ =0\end{array}$

  $\begin{array}{c}\underset{x\to -\infty }{\lim }f\left(x\right)=\underset{x\to -\infty }{\lim }{\left(0.5\right)}^x\\ =\underset{x\to -\infty }{\lim }{\left(\frac{1}{2}\right)}^x\\ =\infty \end{array}$