Q. 97 - Calculus

Question

Use the definition of two-sided and one-sided derivatives, together with properties of limits, to prove that $f\text{'}\left(c\right)$ exists if and only if $f\text{'}\_\left(c\right)$ and $f{\text{'}}_{+}\left(c\right)$ exist and are equal. 

Step-by-Step Solution

Answer

$f_{-}^{\mathrm{\prime }}\left(c\right)\text{ and }{f}_{+}^{\mathrm{\prime }}\left(c\right)\text{ are exist and they are equal. Thus }{f}^{\mathrm{\prime }}\left(c\right)\text{ exist. }$

Step1. Given Information

Consider a function

$f\left(x\right)$

$\text{ Here the objective is to show that }{f}^{\mathrm{\prime }}\left(c\right)\text{ exist if and only if }{f}_{-}^{\mathrm{\prime }}\left(c\right)\text{ and }{f}_{+}^{\mathrm{\prime }}\left(c\right)\text{ exist and are }$equal.

Step2. Left Continuity

Now,

$\begin{array}{r}\underset{x\to c^{-}}{lim} \frac{f\left(x\right)-f\left(c\right)}{x-c}=f_{-}^{\mathrm{\prime }}\left(c\right)\\ \frac{\underset{x\to c^{-}}{lim} f\left(x\right)-f\left(c\right)}{\underset{x\to c^{-}}{lim} x-c}=f_{-}^{\mathrm{\prime }}\left(c\right)\\ \underset{x\to c^{-}}{lim} f\left(x\right)-f\left(c\right)=f_{-}^{\mathrm{\prime }}\left(c\right)\left[\underset{x\to c^{-}}{lim} x-c\right]\\ \underset{x\to c^{-}}{lim} f\left(x\right)-\underset{x\to c^{-}}{lim} f\left(c\right)=0\\ \underset{x\to c^{-}}{lim} f\left(x\right)=f\left(c\right)\end{array}$

Therefore the function is left continuous at $x=c$.

Step3. Right Continuity

Again,

$\begin{array}{r}\underset{x\to c^{+}}{lim} \frac{f\left(x\right)-f\left(c\right)}{x-c}=f_{+}^{\mathrm{\prime }}\left(c\right)\\ \frac{\underset{x\to c^{+}}{lim} f\left(x\right)-f\left(c\right)}{\underset{x\to c^{+}}{lim} x-c}=f_{+}^{\mathrm{\prime }}\left(c\right)\\ \underset{x\to c^{+}}{lim} f\left(x\right)-f\left(c\right)=f_{+}^{\mathrm{\prime }}\left(c\right)\left[\underset{x\to c^{+}}{lim} x-c\right]\\ \underset{x\to c^{+}}{lim} f\left(x\right)-\underset{x\to c^{+}}{lim} f\left(c\right)=0\\ \underset{x\to c^{+}}{lim} f\left(x\right)=f\left(c\right)\end{array}$

 Therefore the function is right continuous at $x=c$.

$\text{ Therefore }{f}_{-}^{\mathrm{\prime }}\left(c\right)\text{ and }{f}_{+}^{\mathrm{\prime }}\left(c\right)\text{ are exist and they are equal. Thus }{f}^{\mathrm{\prime }}\left(c\right)\text{ exist. }$