Chapter 3, Problem 7PS

  $\left(a\right)$

To determine

To graph: The function $y_1=1+\frac{x}{1!}$ and compare with the function $y=e^x$ .

Explanation of Solution

Given information:

The function $y_1=1+\frac{x}{1!}$ .

Graph:

Sketch the graph of the $y_1=1+\frac{x}{1!}$ and $y=e^x$ on the same graphing window.

  

The red curve represents the function $y_1=1+\frac{x}{1!}$ and blue curve represents the function $y=e^x$ .

Interpretation:

The function $y_1=1+\frac{x}{1!}$ is rewritten as $y_1=1+x$ because $1!=1$ . It is observed from the graph that both the functions intersect at $\left(0,1\right)$ . The function $y=e^x$ increases rapidly when compared to $y_1=1+\frac{x}{1!}$ .

  $\left(b\right)$

To determine

To graph: The function $y_2=1+\frac{x}{1!}+\frac{x^2}{2!}$ and compare with the function $y=e^x$ .

Explanation of Solution

Given information:

The function $y_2=1+\frac{x}{1!}+\frac{x^2}{2!}$ .

Graph:

Sketch the graph of the $y_2=1+\frac{x}{1!}+\frac{x^2}{2!}$ and $y=e^x$ on the same graphing window.

  

The red curve represents the function $y_2=1+\frac{x}{1!}+\frac{x^2}{2!}$ and blue curve represents the function $y=e^x$ .

Interpretation:

The function $y_2=1+\frac{x}{1!}+\frac{x^2}{2!}$ is rewritten as $y_2=1+x+\frac{x^2}{2}$ because $1!=1\text{ and }2!=2$ . It is observed from the graph that both the functions intersect each other. The function $y=e^x$ increases rapidly when compared to $y_2=1+\frac{x}{1!}+\frac{x^2}{2!}$ .

  $\left(c\right)$

To determine

To graph: The function $y_3=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}$ and compare with the function $y=e^x$ .

Explanation of Solution

Given information:

The function $y_3=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}$ .

Graph:

Sketch the graph of the $y_3=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}$ and $y=e^x$ on the same graphing window.

  

The red curve represents the function $y_3=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}$ and blue curve represents the function $y=e^x$ .

Interpretation:

The function $y_3=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}$ is rewritten as $y_3=1+x+\frac{x^2}{2}+\frac{x^3}{6}$ because $1!=1,2!=2\text{ and }3!=6$ . It is observed from the graph that both the functions intersect each other. The function $y=e^x$ increases rapidly when compared to $y_3=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}$ .