How to solve this? Determine the continuity and derivability domain for #f(x)# #f:[0,oo)->RR,f(x)=|x-1|sqrtx #

1 Answer
Mar 29, 2017

Answer:

#f(x) = abs(x-1)sqrtx#

is continuous in #[0,+oo)# and differentiable in #[0,1) uu (1,+oo)#

Explanation:

We have that:

#x<1 => x-1 < 0 => abs(x-1) = -(x-1)#

#x > 1 => x-1 > 0 => abs(x-1) = (x-1)#

so:

  • For #x in [0,1)#, #f(x) = (-x+1)sqrt(x)#
  • For #x = 1#, #f(x) = 0#
  • For #x in (1,+oo)#, #f(x) = (x-1)sqrt(x)#

We can deduce that the function is continuous in #[0,1) uu (1,+oo)# since it is the composition of continuous functions.

For #x=1# we have:

#lim_(x->1^-) f(x) = lim_(x->1^-) (-x+1)sqrt(x) = 0#

#lim_(x->1^+) f(x) = lim_(x->1^+) (x-1)sqrt(x) = 0#

which implies:

#lim_(x->1) f(x) = 0 = f(1)#

so the function is continuous also in #x=1#.

Differentiating #f(x)# we have:

  • For #x in [0,1)#, #f'(x) = d/dx((-x+1)sqrt(x)) = -sqrtx -(x+1)/(2sqrt(x)) = (1-3x)/(2sqrtx)#
  • For #x in (1,+oo)#, #f'(x) = d/dx((x-1)sqrt(x)) = sqrt(x) + (x-1)/(2sqrtx) =(3x-1)/(2sqrtx)#

so for #x=1#:

#lim_(x->1^-) f'(x) = lim_(x->1^-) (1-3x)/(2sqrtx) =- 1#

#lim_(x->1^+) f'(x) = lim_(x->1^+) (3x-1)/(2sqrtx) = 1#

As the derivative is not continuous in #x=1# the function #f(x)# is not differentiable.