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

Jun 18, 2017

$f \left(x\right)$ has a vertical asymptote $x = 0$ and no other linear asymptotes.

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

#### Explanation:

Given:

$f \left(x\right) = \sqrt{x} + \frac{1}{\sqrt{x}}$

Note that:

• $\sqrt{x}$ only takes real values when $x \ge 0$.

• $\frac{1}{\sqrt{0}} = \frac{1}{0}$ is undefined.

So we find that $f \left(x\right)$ is well defined precisely when $x > 0$.

As $x \to {0}^{+}$:

• $\sqrt{x} \to 0$

• $\frac{1}{\sqrt{x}} \to \infty$

So $f \left(x\right)$ has a vertical asymptote $x = 0$.

For any $x > 0$ both $\sqrt{x} > 0$ and $\frac{1}{\sqrt{x}} > 0$, so $f \left(x\right)$ has no intercepts with the $x$-axis.

As $x \to \infty$:

• $\sqrt{x} \to \infty$

• $\frac{1}{\sqrt{x}} \to 0$

So $f \left(x\right)$ is asymptotic to the function $y = \sqrt{x}$. It has no horizontal or oblique (slant) asymptotes.

By symmetry, the minimum value of $f \left(x\right)$ occurs where $\sqrt{x} = \frac{1}{\sqrt{x}}$, i.e. where $x = 1$.

$f \left(1\right) = \sqrt{1} + \frac{1}{\sqrt{1}} = 1 + \frac{1}{1} = 2$

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