using f'(x) as the limit of h approaches 0 [f(x+h)-f(x)]/h
find: f'(x), given f(x)=|x|
the graph of f(x) = |x|
is a V with the vertex at the origin.
(made up of y = x in quadrant I and y = -x in quadrant II)
your two graphs are y = x and y = -x
I will do the second part as x ---> 0 for y = -x
= lim (-(x+h) - (-x))/h as x ---> 0
= lim (-x - h + x)/h as x--> 0
= lim -h/h
= -1
do the same with f(x) = y = x to show
a result of +1