Question

g(1-x^2)/f(x+1) Using the rational function [(1-x^2)/(x+1)], draw a graph and answer the following: a) what are the zeroes?b) are there any asymptotes? c) what is the domain and range for this function?d) it it a continuous function?e) are there any values of y= g(x)/f(x) that are undefined? Explain

Answer 1

Given the function

`h(x)=(g(x))/(f(x))=(1-x^2)/(x+1)`

Notice that

`\Rightarrow h(x)=((1-x)(1+x))/(x+1)=1-x,\text{ if x+1}\\e0`

Then, h(x) has a removable discontinuity at x=-1

a) To find the zeroes of the function solve h(x)=0, as shown below,

`\begin{gathered} h(x)=0 \\ \Rightarrow(1-x^2)/(x+1)=0 \\ \Rightarrow1-x^2=0,x+1\\e0 \\ \Rightarrow x^2=1,x+1\\e0 \\ \Rightarrow x=\pm1,,x+1\\e0 \\ \Rightarrow x=1 \end{gathered}`

Thus, the zero of the function is at x=1.

b) There are no asymptotes, only a removable discontinuity as shown above.

c) The domain of h(x) consists of the values of x such that the denominator is different than zero

`\begin{gathered} x+1=0 \\ \Rightarrow x=-1 \\ \Rightarrow\text{domain(h(x))}=\mleft\lbrace x\in\R|x\\e-1\mright\rbrace=(-\infty,-1)\cup(-1,\infty) \end{gathered}`

As for the range of the function, as we established before,

`h(x)=1-x,x\\e-1`

And the range of 1-x is the set of real numbers. Therefore, the range of the function is

`\begin{gathered} \text{range(h(x))}=\mleft\lbrace y\in\R|h(-1)\\e y\mright\rbrace \\ \Rightarrow range(h(x))=\mleft\lbrace y\in\R|2\\e y\mright\rbrace=(-\infty,2)\cup(2,\infty) \end{gathered}`

d) There is a discontinuity at x=-1, although it is 'removable'.

e) Except for the point x=-1, h(x) is well defined for any other value of x. The answer to part e) is x=-1, y=2. (-1,2) is the only point that is not included in the domain nor the range of the rational function