Let [f(x)=cases3x, & x2cases] where [] denotes the great…

Question

Let
$$f(x)=\begin{cases}3x, & x<0\ \min\{1+x+[x],\ x+2[x]\}, & 0\leq x\leq2\ 5, & x>2\end{cases}$$
where $[\cdot]$ denotes the greatest integer function. If $\alpha$ and $\beta$ are the number of points where $f$ is not continuous and not differentiable, respectively, then $\alpha+\beta$ equals \_\_\_\_\_\_\_.

Answer: 5

Accepted Answer

On $0\le x<1$, $[x]=0$: $\min\{1+x,\,x\}=x$. On $1\le x<2$, $[x]=1$: $\min\{2+x,\,x+2\}=x+2$. At $x=2$: $\min\{1+2+2,\ 2+4\}=\min\{5,6\}=5$.

$$f(x)=\begin{cases}3x, & x<0\ x, & 0\le x<1\ x+2, & 1\le x<2\ 5, & x\ge2\end{cases}$$

Continuity: at $x=0$: both pieces give $0$ --- continuous. At $x=1$: $f(1^-)=1$, $f(1)=3$ --- discontinuous. At $x=2$: $f(2^-)=4$, $f(2)=5$ --- discontinuous. So $\alpha=2$ (points $x=1,2$).

Differentiability: at $x=0$, slopes $3$ (left) vs $1$ (right) differ --- non-differentiable. $x=1,2$ are already discontinuous, hence non-differentiable. So $\beta=3$ (points $x=0,1,2$).

$$\alpha+\beta=2+3=5$$

Solutions · 1