# How do you graph #y =(x^(1/2) +1)/(x+1)#?

##### 1 Answer

Analyse this function to find the domain, some points through which it goes, maximum and asymptotic behaviour.

#### Explanation:

Let

Hence the domain of

#f(0) = f(1) = 1#

#f'(x) = (-x^(1/2)+x^(-1/2)-2)/(2(x+1)^2)#

Let

Then

Multiply this equation through by

#t^2+2t-1 = 0#

This has roots:

Since

So

When

#f(x) = (sqrt(x)+1)/(x+1) = sqrt(2)/(4 - 2sqrt(2)) = (sqrt(2)(2+sqrt(2)))/(2(2-sqrt(2))(2+sqrt(2)))#

#=(sqrt(2)+1)/(4-2) = (sqrt(2)+1)/2 ~~ 1.2071#

As

Putting this together, we have a curve only defined for

graph{(x^(1/2)+1)/(x+1) [-0.552, 4.448, -0.78, 1.72]}