Chapter 2.2, Problem 46E

(a)

To determine

To find:A simple function as a right end behavior model for the function $y=x^2+e^{-x}$ .

Answer to Problem 46E

A simple function as a right end behavior model for the function $y=x^2+e^{-x}$ is $x^2$ .

Explanation of Solution

Given information:

The function is $y=x^2+e^{-x}$ .

Calculation:

A function $f\left(x\right)$ is said to be right end behavior model for the function $g\left(x\right)$ if $\underset{x\to \infty }{\lim }\frac{g\left(x\right)}{f\left(x\right)}=1$ .

As $x\to \infty$ , then the function $x^2$ approaches to infinity. So, the term $x^2$ dominates on the right.

The function $f\left(x\right)=x^2$ will be the right-end behavior model.

Verify:

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

Therefore, a simple function as a right end behavior model for the function $y=x^2+e^{-x}$ is $x^2$ .

(b)

To determine

To find:A simple function as a left end behavior model for the function $y=2x^2+e^{-x}$ .

Answer to Problem 46E

A simple function as a left end behavior model for the function $y=2x^2+e^{-x}$ is $e^{-x}$ .

Explanation of Solution

Given information:

The function is $y=2x^2+e^{-x}$ .

Calculation:

A function $f\left(x\right)$ is said to be left end behavior model for the function $g\left(x\right)$ if $\underset{x\to -\infty }{\lim }\frac{g\left(x\right)}{f\left(x\right)}=1$ .

As $x\to -\infty$ , then the function $e^{-x}$ increases and it tends much away from the x -axis. So, the term $e^{-x}$ dominates on the left.

The function $f\left(x\right)=e^{-x}$ is the left end behavior model.

Verify:

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

Therefore, a simple function as a left end behavior model for the function $y=2x^2+e^{-x}$ is $e^{-x}$ .