Question
Consider the function \[f\left( x \right) = \left\{ \begin{array}{l}
\,\,\,\,{x^2},\,\,\,\,\,\,\,\,x > 2\\
3x - 2,\,\,x \le 2\,
\end{array} \right.\]
Which one of the following statements is correct in respect of the above function ?
A.
$$f\left( x \right)$$ is derivable but not continuous at $$x = 2.$$
B.
$$f\left( x \right)$$ is continuous but not derivable at $$x = 2.$$
C.
$$f\left( x \right)$$ is neither continuous nor derivable at $$x = 2.$$
D.
$$f\left( x \right)$$ is continuous as well as derivable at $$x = 2.$$
Answer :
$$f\left( x \right)$$ is continuous but not derivable at $$x = 2.$$
Solution :
$$\eqalign{
& {\text{First we check continuity at }}x = 2 \cr
& {\text{L}}{\text{.H}}{\text{.L}}{\text{.}} = \mathop {\lim }\limits_{h \to 0} f\left( {2 - h} \right) \cr
& = \mathop {\lim }\limits_{h \to 0} 3\left( {2 - h} \right) - 2 \cr
& = \mathop {\lim }\limits_{h \to 0} 4 - 3h \cr
& = 4 \cr
& {\text{R}}{\text{.H}}{\text{.L}}{\text{.}} = \mathop {\lim }\limits_{h \to 0} f\left( {2 + h} \right) \cr
& = \mathop {\lim }\limits_{h \to 0} {\left( {2 + h} \right)^2} \cr
& = 4 \cr
& {\text{Also, }}f\left( 2 \right) = {\left( 2 \right)^2} = 4 \cr
& {\text{Since, L}}{\text{.H}}{\text{.L}}{\text{.}} = {\text{R}}{\text{.H}}{\text{.L}}{\text{.}} = f\left( 2 \right) \cr
& \therefore \,f\left( x \right)\,{\text{is continuous at }}2 \cr
& {\text{Now, we check for differentiability}} \cr
& {\text{L}}{\text{.H}}{\text{.D}}{\text{. at }}x = 2 \cr
& f'\left( x \right) = 3x - 2 \cr
& f'\left( x \right) = 3 \cr
& {\left. {f'\left( x \right)} \right|_{x = 2}} = 3 \cr
& {\text{R}}{\text{.H}}{\text{.D}}{\text{. at }}x = 2 \cr
& f'\left( x \right) = {x^2} \cr
& f'\left( x \right) = 2x \cr
& {\left. {f'\left( x \right)} \right|_{x = 2}} = 4 \cr
& {\text{Since L}}{\text{.H}}{\text{.D}}{\text{.}} \ne {\text{R}}{\text{.H}}{\text{.D}}{\text{.}} \cr
& \therefore \,f\left( x \right){\text{ is not dervable at }}x = 2 \cr} $$