Mathematics - Continuity - Differentiability Question with Solution | TestHub

We analyze $f(x)$ based on $|x|$. If $|x|<1$, $x^{2n}\to 0$, so $f(x)=[x{]}^2$. For $x\in (-1,0)$, $f(x)=1$; for $x\in [0,1)$, $f(x)=0$. If $|x|>1$, $x^{2n}$ dominates, so $f(x)=1$. For $x=1$, $f(1)=\frac{1+[1{]}^2}{1+1}=1$. For $x=-1$, $f(-1)=\frac{1+[-1{]}^2}{1+1}=1$. Checking continuity: At $x=0$, $f(0^{-})=1$, $f(0^{+})=0$, so discontinuous. At $x=1$, $f(1^{-})=0$, $f(1^{+})=1$, so discontinuous. At $x=-1$, $f(-1^{-})=1$, $f(-1^{+})=1$, $f(-1)=1$, so continuous. Thus, $f(x)$ is discontinuous at $x=0$ and $x=1$, making 2 points of discontinuity.