Question

Find whether the function is differentiable at x = 1 and x = 2
$f\left(x\right)=\left\{\begin{array}{ll}x& x\le 1\\ \begin{array}{c}2-x\\ -2+3x-x^2\end{array}& \begin{array}{c}1\le x\le 2\\ x>2\end{array}\end{array}\right.$

Answer 1

$$\begin{gathered} f\left(x\right)=\left\{\begin{array}{ll}x& x\le 1\\ \begin{array}{c}2-x\\ -2+3x-x^2\end{array}& \begin{array}{c}1\le x\le 2\\ x>2\end{array}\end{array}\right. \\ \Rightarrow f\text{'}\left(x\right)=\left\{\begin{array}{ll}1& x\le 1\\ \begin{array}{c}-1\\ 3-2x\end{array}& \begin{array}{c}1\le x\le 2\\ x>2\end{array}\end{array}\right. \end{gathered}$$
$$\begin{gathered} \mathrm{Now}, \\ \mathrm{LHL}=\underset{x\to 1^{-}}{\lim }f\text{'}\left(x\right)=\underset{x\to 1^{-}}{\lim }1=1 \\ \mathrm{RHL}=\underset{x\to 1^{+}}{\lim }f\text{'}\left(x\right)=\underset{x\to 1^{+}}{\lim }-1=-1 \\ \mathrm{Since},\mathrm{at}x=1,\mathrm{LHL}\ne \mathrm{RHL} \\ \mathrm{Hence},f\left(x\right)\mathrm{is}\mathrm{not}\mathrm{differentiable}\mathrm{a}tx=1 \\ \mathrm{Again}, \\ \mathrm{LHL}=\underset{x\to 2^{-}}{\lim }f\text{'}\left(x\right)=\underset{x\to 2^{-}}{\lim }-1=-1 \\ \mathrm{RHL}=\underset{x\to 2^{+}}{\lim }f\text{'}\left(x\right)=\underset{x\to 2^{+}}{\lim }3-2x=3-4=-1 \\ \mathrm{Since},\mathrm{at}x=2,\mathrm{LHL}=\mathrm{RHL} \\ \mathrm{Hence},f\left(x\right)\mathrm{is}\mathrm{differentiable}\mathrm{a}tx=2 \end{gathered}$$