Question 143035
Prove that f(x) = |x-1| is not differentiable at (1,0).
Proof: y=f(x) = |x-1|
We use a direct proof.
{{{((delta)y)/((delta)x)=((delta)f)/((delta) x)}}}=y'={{{lim( h->0,((abs(x-1)+h)-abs(x-1))/h)}}} (first principle)
={{{lim( h->0,((abs(x-1)+h)-abs(x-1))((abs(x-1)+h)+abs(x-1))/
h((abs(x-1)+h)+abs(x-1)))}}} (rationalization)
={{{lim( h->0,((x-1+h)^2-(x-1)^2)/
h((abs(x-1)+h)+abs(x-1)))}}}
={{{lim( h->0,(-2h +h^2+2hx)/
h((abs(x-1)+h)+abs(x-1)))}}}
={{{lim( h->0,(-2+h+2x)/((abs(x-1)+h)+abs(x-1)))}}}
={{{(2x-2)/2abs(x-1)}}}={{{(x-1)/abs(x-1)}}}
When x=1,y'=0/0 which is undefined.
f(x) = |x-1| is not differentiable at x=1.

Q.E.D.