mathslover (mathslover):

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.

4 years ago
OpenStudy (anonymous):

I think f(x) is differentiable at 2 but not differentiable at 1

4 years ago
OpenStudy (anonymous):

at 1 f(x) is discontinuous i think

4 years ago
mathslover (mathslover):

I seem to get it now.. was an easy one. THanks!

4 years ago
mathslover (mathslover):

Confused again !

4 years ago
mathslover (mathslover):

@ganeshie8 : can you guide me from start?

4 years ago
ganeshie8 (ganeshie8):

|dw:1402386676928:dw|

4 years ago
ganeshie8 (ganeshie8):

Clearly at x = 1 the function is discontinuous so its not differentiable

4 years ago
ganeshie8 (ganeshie8):

at x = 2, f(x) is continuous but its not differentiable (why ?)

4 years ago
mathslover (mathslover):

how is it discontinuous ? ( at x =1 )

4 years ago
ganeshie8 (ganeshie8):

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

4 years ago
ganeshie8 (ganeshie8):

the function jumps from 0 to 1 at x = 1

4 years ago
mathslover (mathslover):

OH ... okay! I got that part.

4 years ago
ganeshie8 (ganeshie8):

its called jump discontinuity

4 years ago
ganeshie8 (ganeshie8):

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

4 years ago
mathslover (mathslover):

|x| = - x if x < 0 |x| = x if x greater than or equal to 0

4 years ago
mathslover (mathslover):

$$\lim_ {x\rightarrow 0^{-} } |x| = \lim_{x\rightarrow 0^{-}} -x = 0$$ and : $$\lim_{x\rightarrow 0^{+}} |x| - \lim_{x\rightarrow 0^+ } x = 0$$

4 years ago
mathslover (mathslover):

Not sure for why it is not differentiable..

4 years ago
ganeshie8 (ganeshie8):

left and right side limits being equal to same value tells us that the function is continuous at 0

4 years ago
mathslover (mathslover):

Oh okay got it.. :-)

4 years ago
mathslover (mathslover):

LHD is not equal to RHD .. so, its not differentiable

4 years ago
mathslover (mathslover):

*LHL and RHL

4 years ago
ganeshie8 (ganeshie8):

But for differentiability, we need to consider the slopes on left side and right side

4 years ago
ganeshie8 (ganeshie8):

take the LHL and RHL of slope, and show that they're not equal

4 years ago
ganeshie8 (ganeshie8):

slope = f'(x) = $$\large \lim \limits_{h\to 0} \dfrac{f(x+h)-f(x)}{h}$$

4 years ago
ganeshie8 (ganeshie8):

take the LHL and RHL of above ^

4 years ago
ganeshie8 (ganeshie8):

$$\large \lim \limits_{h\to 0^{-}} \dfrac{f(x+h)-f(x)}{h} = ?$$ $$\large \lim \limits_{h\to 0^{+}} \dfrac{f(x+h)-f(x)}{h} = ?$$

4 years ago
mathslover (mathslover):

h tends to 0 + , then it is 1. when h tends to 0- , then it is -1. Right?

4 years ago
ganeshie8 (ganeshie8):

Exactly !! from the graph also it should be clear

4 years ago
ganeshie8 (ganeshie8):

at x = 2, the graph has two different tangents

4 years ago
ganeshie8 (ganeshie8):

|dw:1402387794762:dw|

4 years ago