Chapter 7.5, Problem 9QR

To determine

To find horizontal asymptotes of the graph of $f$ .

Answer to Problem 9QR

  $y=0$ and $y=60$ are the horizontal asymptotes to $f\left(x\right)=\frac{60}{1+5e^{-0.1x}}$ .

Explanation of Solution

Given information:

The function is $f\left(x\right)=\frac{60}{1+5e^{-0.1x}}$

Calculation:

Horizontal asymptotes are the horizontal lines that the graph of function approaches as $x$ tends to $+\infty$ or $-\infty$ .

Graph of $f\left(x\right)=\frac{60}{1+5e^{-0.1x}}$ is given below

  

From the graph of $f\left(x\right)$ it can be observed that as $x$ tends to $-\infty$ the graph of $f\left(x\right)$ approaches to $y=0$ and as $x$ tends to $+\infty$ the graph of $f\left(x\right)$ approaches to $y=60$ .

Therefore, $y=0$ and $y=60$ are the horizontal asymptotes to $f\left(x\right)=\frac{60}{1+5e^{-0.1x}}$ .