Login

show that f(x) =|x-2| is continuous but not differentiable at x=2. Please reply!!URGENT!!!

a).|x-2| being a modulus function is continuous 

b)For cheking differentiability, at x=2, we need to check if lim h=>0+ ( f(2+h) -f(2)/h) should not be equal to lim h=>0- ( f(2+h) -f(2)/h) or not

Now, lim h=>0+ f(2+h) -f(2)/h) = lim h=>0+ ( !2+h-2! -!2-2! ) /h = lim h=>0+ !h! /h = lim h=>0+ h/h =1 (as h= +ve hence !h!=h)

 

Similarly, lim h=>0- f(2+h) -f(2)/h) = lim h=>0- ( !2+h-2! -!2-2! ) /h = lim h=>0- !h! /h = lim h=>0- - h/h =-1 (as h= -ve hence !h!= -h)

 

so lim h=>0+ f(2+h) -f(2)/h) is  equal to lim h=>0- f(2+h) -f(2)) /h

  • -8
View Full Answer

f(x) = |x+2|

f(x) = x+2 ; x= -2

-(x+2) ; x,-2

so, now,

as it is a polynomial function for all x belongs to R {-2},

therefore, it is continuous as well as differentiable over there.

at -2,

LHL = lt x- (-2)- f(x) = 0

RHL = lt x- (-2)+ f(x) = 0

LHL = RHL

so it is continuous everywhere.

now,

it is differentiable also for all x belongs to R {-2},

therefore now we need to check only at -2.

LHD = d/dx ( -x-2) = -1

RHD = d/dx (x+2) = 1

= LHD ≠ RHD

hence, f(x) is continuous everwhere but not differentiable at x = -2.

you can even see through graph that it is cont. everywhere but there is a kink t -2 indicating it is non-differentiable over there.

hope it helps!!

cheers!!!

  • 2
Here , using the meaning of modulus , we can write the function as
f(x) =     { x – 2 , if x ≥ 2
                { -(x – 2) = 2 – x , if x < 2
f(2) = |2 – 2| = 0
Now  at x = 2,
LHL = lim ( x->2-) f(x) = lim ( x -> 2) ( 2 – x) = 0
RHL – lim( x – 2+) f(x) = lim( x -> 2) ( x – 2) = 0
LHL = RHL = f(2)
SO f(x) is continuous at x = 2.
Now on differentiating the function , we get
Left Hand Derivative ( LHD) = d/dx(  2 – x )  = -1
Right Hand Derivative ( RHD) = d/dx(x – 2) = 1
As LHD ≠ RHS , f(x) is not differentiable at x = 2 .
Hope that helps.
 
  • 41
Now, lim h=>0+ f(2+h) -f(2)/h) = lim h=>0+ ( !2+h-2! -!2-2! ) /h = lim h=>0+ !h! /h = lim h=>0+ h/h =1 (as h= +ve hence !h!=h)
 
Similarly, lim h=>0- f(2+h) -f(2)/h) = lim h=>0- ( !2+h-2! -!2-2! ) /h = lim h=>0- !h! /h = lim h=>0- - h/h =-1 (as h= -ve hence !h!= -h)
 
so lim h=>0+ f(2+h) -f(2)/h) is  equal to lim h=>0- f(2+h) -f(2)) /h
 
  • 1
What are you looking for?