Chapter 3, Problem 13RE

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

  $\underset{x\to \infty }{\lim }f\left(x\right)=2$

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

Given information:

The functions:

  $f\left(x\right)=e^{4-x}+2$

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:

  $f\left(x\right)=e^{4-x}+2$

Here, it can be observed that the coefficient of x in the given function is $-1$ . This will cause the value of the function to decrease as the value of x will increase.

Hence this function is an exponential decay function.

Now, the end behavior of this function is:

  $\begin{array}{c}\underset{x\to \infty }{\lim }f\left(x\right)=\underset{x\to \infty }{\lim }\left(e^{4-x}+2\right)\\ =\underset{x\to \infty }{\lim }\left(e^{4-x}\right)+\underset{x\to \infty }{\lim }\left(2\right)\\ =0+2\\ =2\end{array}$

  $\begin{array}{c}\underset{x\to -\infty }{\lim }f\left(x\right)=\underset{x\to -\infty }{\lim }\left(e^{4-x}+2\right)\\ =\underset{x\to -\infty }{\lim }\left(e^{4-x}\right)+\underset{x\to -\infty }{\lim }\left(2\right)\\ =\infty +2\\ =\infty \end{array}$