Given the piecewise defined function $f(x)=\left\{ \begin{array}{r@{\quad}rr} -2x^{2}+4 & \hbox{if} & x<1 \\ x^{2}+1 & \hbox{if} & x ≥ 1 \end{array} \right. ,$ determine if $f^\prime$ (1) exists.

Solution We investigate the limit $$\lim\limits_{x\rightarrow 1}\frac{f(x)-f(1)}{x-1}=\lim\limits_{x\rightarrow 1}\frac{f(x)-2}{x-1} \qquad {\color{#0066A7}{\hbox{\(f(1)=1^2+1=2\)}}}$$

If $x < 1$, then $f(x)=-2x^{2}+4$; if $x ≥ 1$, then $f(x)=x^{2}+1$. So, it is necessary to investigate the one-sided limits at 1. $$\begin{eqnarray*} \lim\limits_{x\rightarrow 1^{-}}\frac{f(x)-f(1)}{x-1}&=&\lim\limits_{x \rightarrow 1^{-}}\frac{(-2x^{2}+4)-2}{x-1} = \lim\limits_{x\rightarrow 1^{-}} \frac{-2(x^{2}-1)}{x-1}\\[5pt] &=&-2\lim\limits_{x\rightarrow 1^{-}}\frac{(x-1)(x+1)}{ x-1}=-2\lim\limits_{x\rightarrow 1^{-}}(x+1)=-4\\[5pt] \lim\limits_{x\rightarrow 1^{+}}\frac{f(x)-f(1)}{x-1}&=&\lim\limits_{x \rightarrow 1^{+}}\frac{(x^{2}+1)-2}{x-1} = \lim\limits_{x\rightarrow 1^{+}} \frac{(x-1)(x+1)}{x-1} = \lim\limits_{x\rightarrow 1^{+}}(x+1)=2 \end{eqnarray*}$$ Since the one-sided limits are not equal, $\lim\limits_{x\rightarrow 1} \dfrac{f( x) -f(1) }{x-1}$ does not exist, and so $f^\prime (1)$ does not exist.