Show that y=|x-3|is not differentiable at x=3
y = |x-3|
----> y = x-3 for x>3 or y = -x + 3 for x<3
for y = x-3, dy/dx = 1 , for x > 3
for y = -x+3, dy/dx = -1 , for x < 3
look at the graph of y = |x-3|
http://www.wolframalpha.com/input/?i=plot+y+%3D+%7Cx-3%7C
when x = 3, y = 0 for both parts of the function.
so at x = 3, the line can't make up its mind which slope is has, is it +1 or -1
Such indecisions are not valid in mathematics and we cop out and say it is "undefined"
Notice that if we take points very close to (3,0) we will get definite answers.
e.g. for P = (2.999, .001)
slope to (3,0) = (.001-0)/(2.999-3) = .001/-.001 = -1
for Q at (3.001, 0.001)
slope to (3,0) = (.001-0)/(3.001-3) = .001, .001) = +1