Question

For the function f(x) = `sqrtx+1/sqrtx` what are the intercepts and asymptotes ?

Answer 1

`f(x)` has a vertical asymptote `x=0` and no other linear asymptotes.

It does not intercept the `x` or `y` axes.

Explanation:

Given:

`f(x) = sqrt(x)+1/sqrt(x)`

Note that:

• `sqrt(x)` only takes real values when `x >= 0`.

• `1/sqrt(0) = 1/0` is undefined.

So we find that `f(x)` is well defined precisely when `x > 0`.

As `x->0^+`:

• `sqrt(x)->0`

• `1/sqrt(x)->oo`

So `f(x)` has a vertical asymptote `x=0`.

For any `x > 0` both `sqrt(x) > 0` and `1/sqrt(x) > 0`, so `f(x)` has no intercepts with the `x`-axis.

As `x->oo`:

• `sqrt(x)->oo`

• `1/sqrt(x)->0`

So `f(x)` is asymptotic to the function `y = sqrt(x)`. It has no horizontal or oblique (slant) asymptotes.

By symmetry, the minimum value of `f(x)` occurs where `sqrt(x) = 1/sqrt(x)`, i.e. where `x=1`.

`f(1) = sqrt(1)+1/(sqrt(1)) = 1+1/1 = 2`

We now have enough information to draw the graph of `f(x)`...
graph{(y-sqrt(x)-1/sqrt(x))((x-1)^2+(y-2)^2-0.01) = 0 [-4.13, 15.87, -1.64, 8.36]}