Respuesta :
Given the function
[tex]h(x)=\frac{g(x)}{f(x)}=\frac{1-x^2}{x+1}[/tex]Notice that
[tex]\Rightarrow h(x)=\frac{(1-x)(1+x)}{x+1}=1-x,\text{ if x+1}\ne0[/tex]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,
[tex]\begin{gathered} h(x)=0 \\ \Rightarrow\frac{1-x^2}{x+1}=0 \\ \Rightarrow1-x^2=0,x+1\ne0 \\ \Rightarrow x^2=1,x+1\ne0 \\ \Rightarrow x=\pm1,,x+1\ne0 \\ \Rightarrow x=1 \end{gathered}[/tex]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
[tex]\begin{gathered} x+1=0 \\ \Rightarrow x=-1 \\ \Rightarrow\text{domain(h(x))}=\mleft\lbrace x\in\R|x\ne-1\mright\rbrace=(-\infty,-1)\cup(-1,\infty) \end{gathered}[/tex]As for the range of the function, as we established before,
[tex]h(x)=1-x,x\ne-1[/tex]And the range of 1-x is the set of real numbers. Therefore, the range of the function is
[tex]\begin{gathered} \text{range(h(x))}=\mleft\lbrace y\in\R|h(-1)\ne y\mright\rbrace \\ \Rightarrow range(h(x))=\mleft\lbrace y\in\R|2\ne y\mright\rbrace=(-\infty,2)\cup(2,\infty) \end{gathered}[/tex]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
