Question

Which one of the following functions is continuous everywhere in its domain but has at least one point where it is not differentiable?
A. $f(x) = {x^{\dfrac{1}{3}}}$
B. $f(x) = \dfrac{{|x|}}{x}$
C. $f(x) = {e^{ - x}}$
D. $f(x) = \tan x$

Answer 1

First, we will find if the function is continuous over all the points. We will check for continuity by using the first derivative test. If the function is continuous over all the points, then we will see if the function is differentiable or not. To check for differentiability, the first derivative should be defined for all the values. Hence, after analyzing all the choices, we will select the function which continues everywhere in its domain but has at least one point where it is not differentiable.

Complete step by step solution:
According to the question we need to find the function (from the given choices) that is continuous everywhere in its domain but not differentiable at one or more points
So for option A
$f(x) = {x^{\dfrac{1}{3}}}$
Checking continuity by first derivative test
$\Rightarrow f'(x) = \dfrac{1}{3}{x^{\dfrac{1}{3} - 1}}$
On solving the power of x we get,
$\Rightarrow f'(x) = \dfrac{1}{3}{x^{\dfrac{{ - 2}}{3}}}$
As we have ${x^{ - 1}} = \dfrac{1}{x}$ , so we get,
$\Rightarrow f'(x) = \dfrac{1}{{3 \times {x^{\dfrac{2}{3}}}}}$
By observing the above equation we can say that it is a monotonically increasing function so it is continuous at every point in its domain
But we also observe that it is not defined at $x = 0$
$\Rightarrow$ This function is not differentiable at $x = 0$
For option B
$f(x) = \dfrac{{|x|}}{x}$
By definition, we know that this function is discontinuous at $x = 0$
So, this option is incorrect.
So for option C
$f(x) = {e^{ - x}}$
Checking continuity by first derivative test for $f(x) = {e^{ - x}}$
$f'(x) = - {e^{ - x}}$
Which is monotonically decreasing function
$\Rightarrow$ it is continuous at every point in the domain
Also, we can observe that it’s derivative is defined for every point in the domain
$\Rightarrow$ it is differentiable at every point in the domain
$\Rightarrow$ option C is incorrect
So for option D
$f(x) = \tan x$
We know that $tan\left( x \right)$ is discontinuous at $x = (2k - 1)2\pi$ ​ where $k \in I$
$\Rightarrow$ option D is incorrect.

Hence the final solution is Option A.

Note:
In these types of questions where we need to check for continuity and differentiability, we should always check at points where the denominator becomes 0 to easily find the potential points of differentiability and discontinuity.
We must know the first differentiation test to check if a function is continuous at all points in its domain.