Q. 14 - Calculus

Question

The two-sided symmetric difference approximation for the slope of a tangent line (see Exercise 20 in Section 2.1) can sometimes be misleading. Use a sequence of symmetric difference approximations to estimate the derivative of $f\left(x\right)=\left|x\right|$ at $x=0$. What does your sequence of approximations suggest about $f\left(0\right)$? Does this seem right?

Step-by-Step Solution

Answer

Symmetric differentiation is not correct approach to find the derivatives.

Step 1. Given information

The given function is $f\left(x\right)=\left|x\right|$

We have to check whether the function is differentiable at $x=0$ or not by using symmetric differentiation.

Step 2. Check the differentiability at x = 0 by using symmetric differentiation.

$f\left(x\right)=\left|x\right|$ can be written as

$f\left(x\right)=\left\{\begin{array}{l}x, if x\ge 0\\ -x, if x<0\end{array}\right.$

Consider $h=0.5$,

$$\begin{gathered} f\left(c+h\right)=f\left(0+0.5\right) \\ =0.5 \\ f\left(c-h\right)=f\left(0-0.5\right) \\ =-0.5 \\ f\text{'}\left(0\right)=\frac{f\left(c+h\right)-f\left(c-h\right)}{2h} \\ =\frac{0.5-\left(-0.5\right)}{2\left(0.5\right)} \\ =1 \end{gathered}$$

Now consider $h=-0.5$,

$$\begin{gathered} f\left(c+h\right)=f\left(0-0.5\right) \\ =-0.5 \\ f\left(c-h\right)=f\left(0-\left(-0.5\right)\right) \\ =0.5 \\ f\text{'}\left(0\right)=\frac{f\left(c+h\right)-f\left(c-h\right)}{2h} \\ =\frac{-0.5-0.5}{2\left(-0.5\right)} \\ =1 \end{gathered}$$

Here we can see that left hand derivate and right hand derivative are equal, it shows that the function is differentiable at $x=0$. But $f$ is not differentiable at $x=0$ so symmetric differentiation is not correct approach to find the derivatives.