Question

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

Answer 1

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

Explanation:

Let `f(x) = (x^(1/2)+1)/(x+1)`

`x^(1/2)` only takes Real values for `x >= 0`.

Hence the domain of `f(x)` is `[0, oo)`

`f(0) = f(1) = 1`

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

Let `t = x^(1/2)`

Then `f'(x) = 0` when `-t+1/t-2 = 0`

Multiply this equation through by `-t` to get:

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

This has roots:

`t = (-2+-sqrt(2^2-(4xx1xx-1)))/(2*1) = -1+-sqrt(2)`

Since `t = sqrt(x) >= 0`, `sqrt(x) = -1+sqrt(2)`.

So `x = (sqrt(2)-1)^2 = 3-2sqrt(2) ~~ 0.17157`

When `x = 3 - 2sqrt(2)`:

`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 `x->oo` `f(x) -> 0` since the denominator has higher degree than the numerator, but quite slowly due to the `x^(1/2)` degree.

Putting this together, we have a curve only defined for `x >= 0`, that goes through `(0, 1)` and `(1, 1)`, is asymptotic to `y = 0` and has a maximum at `(3-2sqrt(2), (sqrt(2)+1)/2) ~~ (0.17157, 1.2071)`

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