Question

f(x) = x + 1g(x) = 1 - x^2h(x) = f(x)/g(x)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= f(x)/g(x) that are undefined? Explain

Answer 1

Given the functions:

`\begin{gathered} f(x)=x+1, \\ g(x)=1-x^2. \end{gathered}`

We must answer the question for the function:

`h(x)=(f(x))/(g(x))=(x+1)/(1-x^2)\text{.}`

c) Domain and range

First of all, we must analyze the domain of the function. Because the funcion h(x) is given by the equotient of two functions, f(x) and g(x), the domain of h(x) are all the values of x = 1 for which we have

g(x) ≠ 0. We see that the function g(x) is zero when x = 1 or x = -1. So the domain of the function is:

`Domain=(-\infty,-1)\cup(-1,1)\cup(1,\infty).`

The range of a function is the complete set of all possible resulting values of the dependent variable (y, usually) after we have substituted the domain. Plotting the function, we get the following graph:

From the graph, we see that the function h(x) takes all the values from -∞ to +∞. So we conclude that the range of the function is:

`Range=(-\infty,\infty).`

a) Zeros

The zeros of a function are the values for which we have h(x) = 0. Looking at the function, we see that it seems to be zero at x = -1 but that number is not in the domain of the function. So we conclude that the function has not any zero.

b) Asymptotes

By definition, an asymptote is a straight line that constantly approaches a given curve but does not meet at any infinite distance. Looking again at the graph of the function, we see that the function has one asymptote at:

`x=1.`

d) Continuity

Looking at the domain of the function, we see that the domain is a union of intervals because the function is not defined at the values x = -1 and x = 1. So we conclude that the function is not continuous.

e) Not defined values

As we mentioned in the point of the domain, the function is not defined at the values x = -1 and x = 1. The explanation is the one given in that point.

End behaviour of the function

To find the end behaviour of the function, we can look at the graph, or we can also rewrite the function in the following way:

`h(x)=(x+1)/(1-x^2)=((x+1))/((x+1)\cdot(1-x))\text{.}`

For x ≠ -1 we can cancel the repeated terms in numerator and denominator, and we get:

`h(x)=(1)/((1-x))\text{ for }x\\e-1.`

If we compute the limits of the function as x → -∞ and x → ∞, we get:

`\begin{gathered} \lim _(x\to-\infty)h(x)=\lim _(x\to-\infty)(1)/(1-x)=0, \\ \lim _(x\to+\infty)h(x)=\lim _(x\to+\infty)(1)/(1-x)=0, \end{gathered}`

So the end behaviour of the function is y → 0.

Answers

a) The function has not any zeros

b) x = 1

c) Domain = (-∞,-1) U (-1,1) U (1,∞)

Range = (-∞,∞)

d) The function is not continuous.

e) The function is not defined at x = -1 and x = 1