Q.34 - Calculus

Question

Find the roots, discontinuities, and horizontal and vertical asymptotes of the functions in Exercises 23–34. Support your answers by explicitly computing any relevant limits.

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

Step-by-Step Solution

Answer

The function has no real root.

The function has a discontinuity at $x=0$.

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

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

Step 1. Given Information

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

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

Step 2. Roots of a function

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 $-\frac{\mathrm{\pi }}{2}<{\tan }^{-1}x<\frac{\mathrm{\pi }}{2}$.

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

Step 3. Discontinuities of a function

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

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.

Step 4. Horizontal asymptote of a function

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

When $x\to -\infty$,

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

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

When $x\to \infty$,

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

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

Step 5. Vertical asymptote of a function

The vertical asymptote of a rational function are the zeroes of the denominator of the simplified function.

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

Hence the vertical asymptote is $x=0$

Step 6. Conclusion

The function has no real root.

The function has discontinuity at $x=0$

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

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