Chapter 2.2, Problem 11E

To determine

Is the given function has a corner, cusp, vertical tangent and discontinuity.

Answer to Problem 11E

The function is discontinuous.

Explanation of Solution

Given information:

The function is $y=\left\{\begin{array}{c}{\tan }^{-1}x,x\ne 0\\ 1,x=0\end{array}\right\}$ , the function is not differentiable at $x=0$ .

Concept used:

If a function is continuous then it can have cusp, corner or vertical tangent otherwise it will be discontinuous.

Calculation:

Firstly check the continuity.

Value of limit $\underset{x\to 0^{-}}{\lim }\frac{f\left(x\right)-f\left(0\right)}{x-0}$ is as shown below.

  $\begin{array}{c}\underset{x\to 0^{-}}{\lim }\frac{f\left(x\right)-f\left(0\right)}{x-0}=\underset{x\to 0^{-}}{\lim }\frac{{\tan }^{-1}\left(0\right)-1}{x}\\ =\frac{0-1}{0}\\ =-\infty \end{array}$

Value of limit $\underset{x\to 0^{+}}{\lim }\frac{f\left(x\right)-f\left(0\right)}{x-0}$ is as shown below.

  $\begin{array}{c}\underset{x\to 0^{+}}{\lim }\frac{f\left(x\right)-f\left(0\right)}{x-0}=\underset{x\to 0^{+}}{\lim }\frac{{\tan }^{-1}\left(0\right)-1}{x}\\ =\frac{0-1}{0}\\ =\infty \end{array}$

The value of function at $x=0$ is $1$ .

Since all the three values are not equal to one another.

Conclusion: So, the function is discontinuous.