using the definition of a derivative shouw that f(x)=|x| is not differential at x=0.?

Answer 1

f'(x) = (|x+h|-|x|)/h
f'(x) = (|0+h|-|0|)/h
f'(x) = (|h|-0)/h
f'(x) = |h|/h

assume x<0
find h small enough so that x+h<0

lim(h->0) (-(x+h)-(-x))/h
lim(h->0) (-h/h) = lim(h->0) -1 = -1

assume x>0
find h so that x+h>0

lim(h->0) (x+h-x)/h
lim(h->0) (h/h) = lim(h->0) 1 = 1

f'(x) = 1, x>0
f'(x) = -1, x<0

for x=0,

lim(h->0) (f(0+h)-f(0))/h
lim(h->0) (|h|-0)/h = |h|/h

so f'(0) DOES NOT EXIST because there is a jump discontinuity at x=0.

Answer 2

The definition of a derivative is:

lim h->0 of [f(x+h) - f(x)] / h

Plugging our function into the definition we get

lim h->0 of [ |x - h| - |x| ] / h

We notice that as h gets closer to zero, it will become smaller than x and so there is no need for the absolute values. Therefore, it becomes

lim h->0 of -h/h = -1

But we also have to come from the other side, taking values in the negative and going towards zero. Therefore, when we have (-h/h) the -h becomes positive since h is negative. Thus we ALSO have

lim h-> 0 of -h/h = h/h = 1

And since the results are not equal, the derivative doesn't exist at x=0.

Answer 3

f 'x =1
for all values of x f'x = 1 and at 0 the function does not exist. so no differential at 0.

Answer 4

f(x)=!x!={x,x>=0;-x,x<0}
to examine whether f'(0) exists
lim h>0 f(0+h)-f(0)/h
=limh>0f(h)-f(0)/h
=limh>0!h!-0/h
=lim!h!/h.....................(1)
if h is positive,!h!/h=1
if h is negative !h!/h=-1
therefore limh>0+isnoy
=limh>0-!h!/h
therefore lim h>0!h!/h does not exist
therefore from (1)
limh>0 f(0+h)-f(0)/h does not exist
i.e.f'(0) does not exist
i.e.f'(x) does not exist at x=0
so even though the function f(x)=!x!
is continuous at x=0,it is not differentiable at x=0