91Ó°ÊÓ

Given to determine the roots, discontinuities, and horizontal and vertical asymptotes of the functions below.

$f\left(x\right)=\frac{1}{{\tan }^{-1}x}$

The roots of a function are the values of x where the simplified function value is 0.

Simplifying the function and finding the root:

$$\begin{gathered} f\left(x\right)=0 \\ \frac{1}{{\tan }^{-1}x}=0 \end{gathered}$$

Here the value of the denominator is given by $-\frac{\mathrm{Ï€}}{2}<{\tan }^{-1}x<\frac{\mathrm{Ï€}}{2}$.

Hence the expression can never be 0. So, there are no real roots.

A function has a discontinuity at a point $x=a$where $f\left(a\right)≠\underset{x→a}{\lim }f\left(x\right)$.

The function is undefined when the denominator is 0.

$$\begin{gathered} {\tan }^{-1}x=0 \\ x=\tan \left(0\right) \\ x=0 \end{gathered}$$

Hence the function has a discontinuity at 0.

Horizontal asymptote of a function can be determined by the tangents at the infinity i.e. when x is positive or negative infinity.

When $x→-∞$,

$$\begin{gathered} f\left(x\right)=\underset{x→-∞}{\lim }\frac{1}{{\tan }^{-1}x} \\ f\left(x\right)=\frac{1}{{\tan }^{-1}\left(-∞\right)} \\ f\left(x\right)=\frac{1}{-{\displaystyle \frac{\mathrm{π}}{2}}} \\ f\left(x\right)=-\frac{2}{\mathrm{π}} \end{gathered}$$

Hence one horizontal asymptote is$y=-\frac{2}{\mathrm{Ï€}}$

When $x→∞$,

$$\begin{gathered} f\left(x\right)=\underset{x→∞}{\lim }\frac{1}{{\tan }^{-1}x} \\ f\left(x\right)=\frac{1}{{\tan }^{-1}\left(∞\right)} \\ f\left(x\right)=\frac{1}{{\displaystyle \frac{\mathrm{π}}{2}}} \\ f\left(x\right)=\frac{2}{\mathrm{π}} \end{gathered}$$

Hence another horizontal asymptote is$y=\frac{2}{\mathrm{Ï€}}$

The function has no real root.

The function has discontinuity at $x=0$

The horizontal asymptotes of the function are $y=-\frac{2}{\mathrm{Ï€}},y=\frac{2}{\mathrm{Ï€}}$

The vertical asymptote of the function is$x=0$