Show that the function #|x|# is not differentiable at all points?

Question

Aurora Alvarado · Accepted Answer

graph{|x| [-10, 10, -5, 5]}  For the derivative to exist the limit definition of the derivative must exist, and that limit requires a consistent result as you approach #0# from the left and the right. However,  # lim_(x rarr 0^+) (f(x)-f(0))/(x-0) = 1#
# lim_(x rarr 0^-) (f(x)-f(0))/(x-0) = -1# So as we do not have a consistent result then in general # lim_(x rarr 0) (f(x)-f(0))/(x-0) # is not defined, and thus the derivative at #x=0# does not exist. What you are suggesting is taking an average value, but that approach does not hold up to vigour.

Emilia Aguilar · Answer

undefined To show that the function |x| is not differentiable at all points, we can consider the definition of differentiability. A function is differentiable at a point if and only if its derivative exists at that point. Let's examine the function |x| at a specific point, say x = 0. To determine if the derivative exists at this point, we need to calculate the left-hand and right-hand derivatives separately. For x < 0 (left-hand side): The function |x| can be written as -x for x < 0. Taking the derivative of -x, we get -1. Therefore, the left-hand derivative at x = 0 is -1. For x > 0 (right-hand side): The function |x| can be written as x for x > 0. Taking the derivative of x, we get 1. Therefore, the right-hand derivative at x = 0 is 1. Since the left-hand derivative (-1) is not equal to the right-hand derivative (1), the derivative of |x| does not exist at x = 0. Hence, the function |x| is not differentiable at x = 0. By following a similar approach, we can show that the function |x| is not differentiable at any point where x = 0. Therefore, we can conclude that the function |x| is not differentiable at all points.