#f(x)= |x-1|/e^x# Study the derivability for #f(x)# in #x_"0"=1# . What is the derivate for #f'(1)#?

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Answer 1 · 2017-09-07T23:27:42

The function:

#f(x) = abs(x-1)/e^x#

is differentiable for every #x != 1# and #f'(1)# does not exist.

The function #e^x# is differentiable and non null for every #x in RR#.

The function #abs(x-1)# is differentiable for every #x != 1#.

In fact for #x=1#:

#lim_(h->0^-) (abs(x-1-h) - abs(x-1))/h = lim_(h->0^-) abs(-h) /h = lim_(h->0^-) -h/h = -1#

#lim_(h->0^+) (abs(x-1-h) - abs(x-1))/h = lim_(h->0^+) abs(-h) /h = lim_(h->0^+) h/h = 1#

The function:

#f(x) = abs(x-1)/e^x#

is then differentiable for every #x != 1# and #f'(1)# does not exist.