Chapter 6.1, Problem 1CE

a.

To determine

Find the value of given algebraic function.

Answer to Problem 1CE

At the different value of $x=10,20,30,40,50$ , the solution of the algebraic functions is

  $\begin{array}{l}\text{1}\text{. }\left[\text{0}\text{.0990, 0}\text{.0499, 0}\text{.0333, 0}\text{.0250, 0}\text{.01999}\right]\\ \text{2}\text{. [0}\text{.9901, 0}\text{.9975, 0}\text{.9989, 0}\text{.9994, 0}\text{.9996]}\\ \text{3}\text{. [9}\text{.091, 19}\text{.048, 29}\text{.032, 39}\text{.024, 49}\text{.01961]}\end{array}$

Explanation of Solution

Given: The functions are given

  $\begin{array}{l}1)\frac{x}{x^2+1}\\ 2)\frac{x^2}{x^2+1}\\ 3)\frac{x^2}{x+1}\end{array}$

And the value of x is also given

  $x=10,20,30,40,50$

Calculation:

Let’s take all the values of x, put into the given algebraic functions and find the solution.

Here we have used MATLAB to simplify the functions.

program:

clc
clear
close all
syms x
f=x./(x^2+1);
f1=x^2./(x^2+1);
f2=x^2./(x+1);
a=[10 20 30 40 50];
for i=1:5
    I=subs(f,x,a(i));
    I1=subs(f1,x,a(i));
    I2=subs(f2,x,a(i));
    ans(i)=double(I);
    ans1(i)=double(I1);
    ans2(i)=double(I2);
end

Query:

• First, we have defined the given functions.
• Then define the value of x.
• The using function “subs” in MATLAB to put the values of x into the function and get the solution.
• Similar process we will repeat for all the functions.

b.

To determine

Explain the solution of the algebraic function

Answer to Problem 1CE

At the different value of $x=10,20,30,40,50$ , the solution of the algebraic functions is

  $\begin{array}{l}\text{1}\text{. }\left[\text{0}\text{.0990, 0}\text{.0499, 0}\text{.0333, 0}\text{.0250, 0}\text{.01999}\right]\\ \text{2}\text{. [0}\text{.9901, 0}\text{.9975, 0}\text{.9989, 0}\text{.9994, 0}\text{.9996]}\\ \text{3}\text{. [9}\text{.091, 19}\text{.048, 29}\text{.032, 39}\text{.024, 49}\text{.01961]}\end{array}$

Explanation of Solution

Given: The functions are given

  $\begin{array}{l}1)\frac{x}{x^2+1}\\ 2)\frac{x^2}{x^2+1}\\ 3)\frac{x^2}{x+1}\end{array}$

And the value of x is also given

  $x=10,20,30,40,50$

Calculation:

As we can see in the answers, the solution of first function is exponentially decreasing because the value of $x^2>>x$ .

Second solution exponentially increasing because $x^2+1\approx x^2$ .

In third solution the value of the function continuously increasing because $x^2>>x+1$ .