Question
Determine whether the following function is differentiable at $x=3$ where,
$f(x)=x^2+2, \text { for } x \geq 3$
$=6 x-7, \text { for } x<3$

Answer

$f(x)=x^2+2, x \geq 3$
$=6 x-7, x<3$
Differentiability at $x=3$
$
\begin{aligned}
\operatorname{Lf}^{\prime}(3) & =\lim _{h \rightarrow 0^{-}} \frac{f(3+h)-f(3)}{h} \\
& =\lim _{h \rightarrow 0^{-}} \frac{6(3+h)-7-\left(3^2+2\right)}{h} \\
& =\lim _{h \rightarrow 0^{-}} \frac{18+6 h-7-11}{h} \\
& =\lim _{h \rightarrow 0^{-}} \frac{6 h}{h}
\end{aligned}
$
$=\lim _{h \rightarrow 0^{-}} 6 \quad \ldots[\because h \rightarrow 0, \therefore h \neq 0]$
$=6$
$\operatorname{Rf}^{\prime}(3)=\lim _{h \rightarrow 0^{+}} \frac{f(3+h)-f(3)}{h}$
$=\lim _{h \rightarrow 0^{+}} \frac{(3+h)^2+2-\left(3^2+2\right)}{h}$
$=\lim _{h \rightarrow 0^{+}} \frac{h^2+6 h+9+2-11}{h}$
$=\lim _{h \rightarrow 0^{+}} \frac{h^2+6 h}{h}$
$=\lim _{h \rightarrow 0^{+}}(h+6) \ldots[\because h \rightarrow 0, \therefore h \neq 0]$
$=6$
$\text { Here, Lf }(3)=\text { Rf' }(3)$
$\therefore \text { fis differentiable at } x=3 .$

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