if-f-x-x-3-2-f-0-0-or-not-exist-

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

i f f (x) = x^{\frac{3}{2}}

f^{'} (0) = 0 o r n o t e x i s t

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 p o s t e d t h i s q u e s t i o n f r o m t h i s e x a m

a n d t h a n k s f o r a l l

Answered by frc2crc last updated on 18/Jul/20

\frac{d}{d x} x^{3 / 2} = \frac{3}{2} x^{3 / 2 - 1} = \frac{3}{2} \sqrt{x}

s o f^{'} (0) = 0

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

t h a n k y o u s i r

\underset{x \to 0^{+}}{l i m f} (x) = 0 b u t \underset{x \to 0^{-}}{l i m f} (x) i s n o t e x i s t

s o i t h i n k w e c a n t f i n d f^{'} (0)

Commented by frc2crc last updated on 18/Jul/20

i t {d o e s n}^{'} t m a k e a n y s e n s e

\sqrt{x} i s^{'} {o n l y}^{'} f o r p o s i t i v e

n u m b e r s, w h y d o y o u n e e d

a l i m i t ?

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

i a s k e d a b o u t t h e l i m i t b e c a u s e

f^{'} (a) = \underset{x \to a}{l i m} \frac{f (x) - f (a)}{x - a}

f^{'} (0) = \underset{x \to 0}{l i m} \frac{x \sqrt{x} - 0}{x - 0} = \underset{x \to 0}{l i m} \sqrt{x}

a n d i {w a s n}^{,} t s u r e i f \underset{x \to 0}{l i m} \sqrt{x} = 0

a n d t h a n k y o u f o r y o u r e x p l a i n i n g

Answered by OlafThorendsen last updated on 18/Jul/20

\frac{f (0 + h) - f (0)}{h} = \frac{h^{\frac{3}{2}} - 0}{h} = \sqrt{h}

\lim_{h \to 0^{+}} \frac{f (0 + h) - f (0)}{h} = 0

\lim_{h \to 0^{-}} \frac{f (0 + h) - f (0)}{h} does not exist

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

y e s s i r t h a t w h a t w a s i m e a n

Commented by bubugne last updated on 18/Jul/20

w i t h k = - h

h = i^{2} k (k ⩾ 0)

\lim_{h \to 0^{-}} \sqrt{h} = \lim_{k \to 0^{+}} i \sqrt{k} = i \times 0 = 0

Commented by prakash 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) = \sqrt{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 \to x_{0}} \frac{f (x) - f (x_{0})}{x - x_{0}} exists .

\lim_{x \to 0^{+}} \frac{x^{\frac{3}{2}} - 0}{x - 0} = 0, \lim_{x \to 0^{-}} \frac{x^{\frac{3}{2}} - 0}{x - 0} {doesn}^{'} t exist .

Therefore f (x) is not differentiable at x = 0

Commented by prakash 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) = \sqrt{x} is continuous at x = 0 .

It is incorrect to take LHL for

function \sqrt{x}

Commented by Ar Brandon last updated on 19/Jul/20

OK Sir. Thanks for the clarification.