Answer to Question #200193 in Real Analysis for Rajkumar

Question

Examine the function f : R→R defined by

f(x) = { (1/6) (x+1)³ x≠ 0

f(x) = { 5/6 x=0

for continuity on R. If it is not continuous at any point of R, find the nature of

discontinuity there.

Answer 1

Let us examine the function $f : \R→\R$ defined by

$f(x) = \begin{cases} \frac{1}6 (x+1)^3,\ x≠ 0\\ \frac{5}6,\ \ x=0 \end{cases}$

for continuity on $\R.$

Since the elementary function $g(x)= \frac{1}6 (x+1)^3$ is continuous, we conclude that the function $f$ is continuous on the set $\R\setminus\{0\}.$ Taking into account that

$\lim\limits_{x\to 0} f(x)=\lim\limits_{x\to 0} \frac{1}6 (x+1)^3=\frac{1}6\ne\frac{5}6=f(0),$

we conclude that the function $f$ has removable discontinuity at the point $x=0.$