Question Number 103998 by  M±th+et+s last updated on 18/Jul/20

if    f(x)=x^(3/2)   f′(0)=0  or not exist

Commented by M±th+et+s last updated on 19/Jul/20

Commented by M±th+et+s last updated on 19/Jul/20

Commented by M±th+et+s last updated on 19/Jul/20

After reviewing some books I found this

Commented by M±th+et+s last updated on 19/Jul/20

Commented by M±th+et+s last updated on 19/Jul/20

i posted this question from this exam     and thanks for all

Answered by frc2crc last updated on 18/Jul/20

(d/dx)x^(3/2) =(3/2)x^(3/2−1) =(3/2)(√x)  so f′(0)=0

Commented by M±th+et+s last updated on 18/Jul/20

thank you sir    lim_(x→0^+ ) f(x)=0   but lim_(x→0^− ) f(x) is not exist  so i think we cant find f′(0)

Commented byfrc2crc last updated on 18/Jul/20

it doesn′t make any sense  (√(x ))is ′only′ for positive  numbers,why do you need  a limit?

Commented by M±th+et+s last updated on 18/Jul/20

i asked about the limit because  f′(a)=lim_(x→a) ((f(x)−f(a))/(x−a))  f′(0)=lim_(x→0) ((x(√x)−0)/(x−0))=lim_(x→0) (√x)    and i wasn^, t sure if lim_(x→0) (√x)=0    and thank you for your explaining

Answered by OlafThorendsen last updated on 18/Jul/20

((f(0+h)−f(0))/h) = ((h^(3/2) −0)/h) = (√h)  lim_(h→0^+ ) ((f(0+h)−f(0))/h) = 0  lim_(h→0^− ) ((f(0+h)−f(0))/h) does not exist

Commented by M±th+et+s last updated on 18/Jul/20

yes sir that what was i mean

Commented bybubugne last updated on 18/Jul/20

with k = −h  h = i^2 k (k≥0)  lim_(h→0^− )  (√h) = lim_(k→0^+ )  i (√k) = i × 0 = 0

Commented byprakash jain last updated on 18/Jul/20

If the domain of the function has  left end point then you only need  RHL to prove continuity. You  do not go outside of domain and  say function is discontinuous.  f(x)=(√x) is continuous at x=0.

Answered by Ar Brandon last updated on 18/Jul/20

f(x) is differentiable at a point x_0  iff  lim_(x→x_0 ) ((f(x)−f(x_0 ))/(x−x_0 )) exists.  lim_(x→0^+ ) ((x^(3/2) −0)/(x−0))=0 , lim_(x→0^− ) ((x^(3/2) −0)/(x−0)) doesn′t exist.  Therefore f(x) is not differentiable at x=0

Commented byprakash jain last updated on 18/Jul/20

  If the domain of the function has  left end point then you only need  RHL to prove continuity. You  do not go outside of domain and  say function is discontinuous.  f(x)=(√x) is continuous at x=0.  It is incorrect to take LHL for  function (√x)

Commented byAr Brandon last updated on 19/Jul/20

OK Sir. Thanks for the clarification.