Question

examine the f:r->r defined by f(x)={1/6(x+1)^3 x is not equal to 0 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

$$f(x)=\begin{cases} \dfrac{1}{6}(x+1)^3 , x\not=0\\ \\ \dfrac{5}{6}, x=0 \end{cases}$$

The function $f(x)$ is continuous on $(-\infin, 0)\cup (0, \infin)$ as polynomial.

$$\lim\limits_{x\to0}f(x)=\lim\limits_{x\to0}\dfrac{1}{6}(x+1)^3=\dfrac{1}{6}$$

$$\lim\limits_{x\to0}f(x)=\dfrac{1}{6}\not=\dfrac{5}{6}=f(0)$$

The function $f(x)$ has a removable discontinuity at $x=0.$