In this section, we introduce the concept of a one-sided derivative.
Definition 1. If the function \(y=f(x)\) is defined for \(x=x_{0}\), then the derivative from the right of \(f\) at \(x_{0}\), denoted by \(f’_{+}(x_{0})\), is defined by \[f’_{+}(x_{0})=\lim_{\Delta x\to0^{+}}\frac{f(x_{0}+\Delta x)-f(x_{0})}{\Delta x},\]
if the limit exists. Similarly, the derivative from the left of \(f\) at \(x_{0}\), denoted by \(f’_{-}(x_{0})\), is defined by \[f’_{-}(x_{0})=\lim_{\Delta x\to0^{-}}\frac{f(x_{0}+\Delta x)-f(x_{0})}{\Delta x}.\]
The geometric interpretation of one-sided derivatives is illustrated in Figure 1.
Figure 1. The derivative of \(f\) at \(x_{0}\) from the right \(f’_{+}(x_{0})\) is the limit of the secant line slopes connecting \(P(x_{0},f(x_{0}))\) and \(Q(x,f(x))\) as \(Q\) approaches \(P\) from the right. Similarly \(f’_{-}(x_{0})\) is the limit of the secant line slopes as \(Q\) approaches \(P\) from the left.
The Theorem of the Uniquness of a Limit that we learned in the previous chapter, states that \(\lim_{x\to a}f(x)=L\) if and only if \(\lim_{x\to a^{+}}f(x)=\lim_{x\to a^{-}}f(x)=L\). It follows from this theorem that \(f'(x_{0})\) exists if and only if \(f’_{+}(x_{0})=f’_{-}(x_{0})\).
Example
Let \(f\) be the function defined by \[f(x)=\begin{cases} 3x-1 & \text{if }x<2\\ 7-x & \text{if }2\leq x \end{cases}.\]
Draw a sketch of the graph of \(f\). Show that \(f\) is continuous at \(x=2\), and find the derivative from the right and from the left of \(f\) at \(x=2\).
Solution
The graph of \(f\) consists of two lines as shown in Fig. 2. To show that \(f\) is continuous at \(x=2\), we show that the three following conditions hold true
(i) \(f(2)=7-2=5\).
(ii) \({\displaystyle \lim_{x\to2^{-}}f(x)=\lim_{x\to2^{-}}(3x-1)=6-1=5.}\) \({\displaystyle \lim_{x\to2^{+}}f(x)=\lim_{x\to2^{+}}(7-x)=7-2=5.}\)
Therefore \({\displaystyle \lim_{x\to2}f(x)=5}\)
(iii) \({\displaystyle \lim_{x\to2}f(x)=f(2)=5.}\)
Because (i), (ii), and (iii) hold true, \(f\) is continuous at \(x=2\). \[\begin{aligned} f_{+}'(2) & =\lim_{\Delta x\to0^{+}}\frac{f(2+\Delta x)-f(2)}{\Delta x}\\ & =\lim_{\Delta x\to0^{+}}\frac{[7-(2+\Delta x)]-5}{\Delta x}\\ & =\lim_{\Delta x\to0^{+}}\frac{-\Delta x}{\Delta x}\\ & =-1.\end{aligned}\] \[\begin{aligned} f_{-}'(2) & =\lim_{\Delta x\to0^{-}}\frac{f(2+\Delta x)-f(2)}{\Delta x}\\ & =\lim_{\Delta x\to0^{-}}\frac{[3(2+\Delta x)-1]-5}{\Delta x}\\ & =\lim_{\Delta x\to0^{-}}\frac{3\Delta x}{\Delta x}\\ & =3.\end{aligned}\]
Because \[\lim_{\Delta x\to0^{+}}\frac{f(2+\Delta x)-f(2)}{\Delta x}\neq\lim_{\Delta x\to0^{-}}\frac{f(2+\Delta x)-f(2)}{\Delta x},\]
we conclude that \[f'(2)=\lim_{\Delta x\to0}\frac{f(2+\Delta x)-f(2)}{\Delta x}\] does not exists and the function is not differentiable at \(x=2\). However, the derivative from the right and from the left of \(f\) at \(x=2\), \(f’_{+}(2)\) and \(f’_{-}(2)\), exist.
Figure 2. Graph of the piecewise-defined function \(y=f(x)=\begin{cases} 3x-1 & \text{if }x<2\\ 7-x & \text{if }2\leq x \end{cases}\)
