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

Question

Nathan Axel · Accepted Answer

Horizontal asymptote at #y=0#
Domain: #[0, oo)# 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]}

Ethan Adkins · Answer

undefined To graph and find the discontinuities of the function (x^(1/2) + 1)/(x + 1), follow these steps:

1. Determine the domain of the function by identifying any values of x that would make the denominator zero. In this case, x + 1 cannot be zero, so x ≠ -1. Therefore, the domain is all real numbers except x = -1.

2. To graph the function, start by plotting some points. Choose various values of x, calculate the corresponding y-values using the function, and plot the points on a graph.

3. As x approaches -1 from the left side (x < -1), the function approaches negative infinity. As x approaches -1 from the right side (x > -1), the function approaches positive infinity. This indicates a vertical asymptote at x = -1.

4. Additionally, the function has a horizontal asymptote at y = 0, as the degree of the numerator is less than the degree of the denominator.

5. Plot the points and connect them smoothly, considering the asymptotes. The graph should resemble a curve approaching the vertical asymptote at x = -1 and the horizontal asymptote at y = 0.

6. The function has a discontinuity at x = -1, which is a vertical asymptote. Discontinuities occur when the function is undefined or approaches infinity at a certain point.

In summary, the graph of the function (x^(1/2) + 1)/(x + 1) has a vertical asymptote at x = -1, a horizontal asymptote at y = 0, and a discontinuity at x = -1.