Examine the differentiability of f(x) at x = 1 and x = 2 defined by : f(x) = x[x] ; \(0\le x < 2\) = (x-1)x ; \(2 \le x \le 3\) where [x] is the greatest integer function.
I think f(x) is differentiable at 2 but not differentiable at 1
at 1 f(x) is discontinuous i think
I seem to get it now.. was an easy one. THanks!
Confused again !
@ganeshie8 : can you guide me from start?
|dw:1402386676928:dw|
Clearly at x = 1 the function is discontinuous so its not differentiable
at x = 2, f(x) is continuous but its not differentiable (why ?)
how is it discontinuous ? ( at x =1 )
f(x) = x[x] for 0<x<1, [x] = 0 so, f(x) = x*0 = 0 for 1<x<2, [x] = 1 so, f(x) = x*1 = x
the function jumps from 0 to 1 at x = 1
OH ... okay! I got that part.
its called `jump discontinuity`
good :) for seeing why the function is not differentiable at x = 2, we need to knw why the absolute value function is not differentiable at x = 0
|x| = - x if x < 0 |x| = x if x greater than or equal to 0
\(\lim_ {x\rightarrow 0^{-} } |x| = \lim_{x\rightarrow 0^{-}} -x = 0 \) and : \(\lim_{x\rightarrow 0^{+}} |x| - \lim_{x\rightarrow 0^+ } x = 0\)
Not sure for why it is not differentiable..
left and right side limits being equal to same value tells us that the function is continuous at 0
Oh okay got it.. :-)
LHD is not equal to RHD .. so, its not differentiable
*LHL and RHL
But for differentiability, we need to consider the slopes on left side and right side
take the LHL and RHL of slope, and show that they're not equal
slope = f'(x) = \(\large \lim \limits_{h\to 0} \dfrac{f(x+h)-f(x)}{h}\)
take the LHL and RHL of above ^
\(\large \lim \limits_{h\to 0^{-}} \dfrac{f(x+h)-f(x)}{h} = ?\) \(\large \lim \limits_{h\to 0^{+}} \dfrac{f(x+h)-f(x)}{h} = ?\)
h tends to 0 + , then it is 1. when h tends to 0- , then it is -1. Right?
Exactly !! from the graph also it should be clear
at x = 2, the graph has two different tangents
|dw:1402387794762:dw|
Join our real-time social learning platform and learn together with your friends!