Question

Examine the differentialibilty of the function f defined by
$f\left(x\right)=\left\{\begin{array}{ll}2x+3& \mathrm{if}-3\le x\le -2\\ \begin{array}{c}x+1\\ x+2\end{array}& \begin{array}{c}\mathrm{if}-2\le x<0\\ \mathrm{if}0\le x\le 1\end{array}\end{array}\right.$

Answer 1

$$\begin{gathered} f\left(x\right)=\left\{\begin{array}{ll}2x+3& \mathrm{if}-3\le x\le -2\\ \begin{array}{c}x+1\\ x+2\end{array}& \begin{array}{c}\mathrm{if}-2\le x<0\\ \mathrm{if}0\le x\le 1\end{array}\end{array}\right. \\ \Rightarrow f\text{'}\left(x\right)=\left\{\begin{array}{ll}2& \mathrm{if}-3\le x\le -2\\ \begin{array}{c}1\\ 1\end{array}& \begin{array}{c}\mathrm{if}-2\le x<0\\ \mathrm{if}0\le x\le 1\end{array}\end{array}\right. \end{gathered}$$
$$\begin{gathered} \mathrm{Now}, \\ \mathrm{LHL}=\underset{x\to -2^{-}}{\lim }f\text{'}\left(x\right)=\underset{x\to -2^{-}}{\lim }2=2 \\ \mathrm{RHL}=\underset{x\to -2^{+}}{\lim }f\text{'}\left(x\right)=\underset{x\to -2^{+}}{\lim }1=1 \\ \mathrm{Since},\mathrm{at}x=-2,\mathrm{LHL}\ne \mathrm{RHL} \\ \mathrm{Hence},f\left(x\right)\mathrm{is}\mathrm{not}\mathrm{differentiable}\mathrm{a}tx=-2 \\ \mathrm{Again}, \\ \mathrm{LHL}=\underset{x\to 0^{-}}{\lim }f\text{'}\left(x\right)=\underset{x\to 0^{-}}{\lim }1=1 \\ \mathrm{RHL}=\underset{x\to 0^{+}}{\lim }f\text{'}\left(x\right)=\underset{x\to 0^{+}}{\lim }1=1 \\ \mathrm{Since},\mathrm{at}x=0,\mathrm{LHL}=\mathrm{RHL} \\ \mathrm{Hence},f\left(x\right)\mathrm{is}\mathrm{differentiable}\mathrm{a}tx=0 \end{gathered}$$