Question

How do you graph and find the discontinuities of `(x^(1/2) +1)/(x+1)`?

Answer 1

Horizontal asymptote at `y=0`
Domain: `[0, oo)`

Explanation:

Given: `(x^(1/2)+1)/(x+1) = (sqrt(x)+1)/(x+1)`

Discontinuities are caused by holes, jumps, vertical asymptotes, or specific function limitations.

This equation is a rational function in the form `R(x)=(N(x))/(D(x))`

Holes are found if any factors from the numerator can be canceled from the denominator. In this equation no factors can be canceled, so there are no holes.

There is a vertical asymptote at `x = -1, (D(x)=0)`, but this is not a discontinuity for this function because the square root in the numerator limits the function's domain. The domain doesn't start until `x =0`.

Domain is limited due to `sqrt(x): " "x >= 0`

To graph the function you would want to know that there is a horizontal asymptote.

Rational Functions `R(x)=(N(x))/(D(x)) = (a_nx^n+...)/(b_mx^m+...)`

If `n=m`, the horizontal asymptote is `y = (a_n)/(b_m)`

If `n < m`, the horizontal asymptote is `y = 0`

If `n > m`, there is no horizontal asymptote.

If `n = m+1` there is a slant asymptote.

In the example given, `n=1/2 < m = 1`; horizontal asymptote at `y=0`

To graph the function find the `x`-intercept(s), `N(x)=0`:

`sqrt(x)+1 = 0; sqrt(x) = -1 " "` No `x`-intercepts

Find the `y`-intercept, by letting `x=0: " " y = 1`

Graph:
graph{(sqrt(x)+1)/(x+1) [-4.205, 15.795, -4.76, 5.24]}