why derivative doesn't exits at Corner....?
such as f(x) = | x | this function is continuous at x = 0 even then f'(0) doesn't have derivative why ?

Question

why derivative doesn't exits at Corner....?
such as f(x) = | x |     this function is continuous at x = 0 even then f'(0) doesn't have derivative why ?

amistre64 · Answer

becasue a derivative is defined as the left and right limits

amistre64 · Answer

a corner doesnt have the same slope from the left and right to be defined as a derivative

anonymous · Answer

lim f(0) exists ?
x->0

amistre64 · Answer

as x to 0 from the left matches x to 0 from teh right

amistre64 · Answer

|dw:1335937491957:dw|

amistre64 · Answer

notice that the slope for |x| is not the same fromteh left and right at x=0

therefore it is undefined

anonymous · Answer

in the above function |x|   , lim x -> 0 f(0) exists?

anonymous · Answer

|dw:1335937582707:dw|

amistre64 · Answer

limit from the right = 1

limit from the left = -1

-1 not equal 1 so derivative doesnt exist at x=0

anonymous · Answer

@amistre64   but the graph of function |x| is that i sent u ...... there is a corner at x = 0..

amistre64 · Answer

i am well aware of what the graph of |x| looks like, how it behaves, and what its downfalls are when it comes to taking its derivatives