Question

I’m studying for my math final and I cant seem how to solve this question. As I don’t know where to start. From 2a. - 2c.

Answer 1

a)

To write the expression for G(n) as a single fraction, write all the terms of the expression using a common denominator.

The given expression for G(n) is:

`G(n)=(2)/(n+2)-(4)/(n+1)+1`

The denominator of the first term is n+2. The denominator of the second term is n+1. The denominator of the last term is 1.

The least common multiple of the expressions n+2, n+1 and 1 is:

`(n+1)(n+2)`

Use (n+1)(n+2) as a common denominator. Multiply the numerator and the denominator of each term times the factor needed for the denominator to become equal to (n+1)(n+2):

`\begin{gathered} G(n)=(2)/(n+2)\cdot(n+1)/(n+1)-(4)/(n+1)\cdot(n+2)/(n+2)+1\cdot((n+1)(n+2))/((n+1)(n+2)) \\ =(2(n+1))/((n+1)(n+2))-(4(n+2))/((n+1)(n+2))+((n+1)(n+2))/((n+1)(n+2)) \end{gathered}`

Now all the terms of G(n) have the same denominator. We can add the fractions by adding their numerators:

`\Rightarrow G(n)=(2(n+1)-4(n+2)+(n+1)(n+2))/((n+1)(n+2))`

Expand the products and simplify the expressions in the numerator and the denominator:

`\begin{gathered} \Rightarrow G(n)=(2n+2-4n-8+n^2+3n+2)/(n^2+2n+n+2)_{} \\ \Rightarrow G(n)=(n^2+2n-4n+3n+2-8+2)/(n^2+3n+2)_{} \\ \Rightarrow G(n)=(n^2+n-4)/(n^2+3n+2)_{} \end{gathered}`

Therefore, the formula for G(n) written as a single fraction, is:

`G(n)=(n^2+n-4)/(n^2+3n+2)`

b)

A rational function is a function that can be expressed as the quotient between two polynomials P(x) and Q(x), with Q(x) a nonzero polynomial:

`f(x)=(P(x))/(Q(x))`

As we can see, the numerator and the denominator of G(n) written as a single fraction, are both polynomials. Therefore, G(n) is a rational function.

c)

The leading coefficient of the numerator of G(n) is 1 and the leading coefficient of its denominator is 1. Both the numerator and the denominator of G(n) have the same degree, which is equal to 2. Then, the end behavior of the function G(n) is:

`G(n)\rightarrow(1)/(1)=1`

Then, G(n) approaches 1 as n grows. In the context of the problem, this means that the Gini index for the particular group tends towards 1 when n grows. This means that inequality increases when the size of the group increases.