Is the function differentiable at x=1

Question

The question is f(x) = | x-1 | + | x+1 | is differentiable at x=1 or not .

 

Now ,

when x<1 , the first part becomes : -(x-1) , i.e 1-x

and why should we not change the sign of second part ,i.e (x+1) ?

Answer 1

Let us rewrite the function as :

$f(x) = \begin{cases} -(x-1)-(1+x), & \text{if }x\leq -1\\ -(x-1)+(1+x), & \text{if }-1 < x\leq 1 \\ +(x-1)+(1+x), & \text{if }1 < x \end{cases}$

Hope it clears your doubt. If not, then take a value for each of the three intervals, put in the formula and convince yourself that it is correct.

Comments

hi gaurav/coolcoder,

just a doubt ,

i dont understand the second equation

−(x−1)+(1+x), if −1<x≤1

why a '-' in the left in(x-1) and a '+' in the middle . 

thanks.

Answer 2

Because it it addition? o.O How will you change the sign? from (x+1) to +(1+x)?

we are not changing the sign bcz x-1 is less than 1. we are changing it bcoz say, x=0.9999, then 0.9999-1=-0.00001, which is a -ve value. after coming out of mod, it will be +ve.

Comments

See , I know that.. It is addition so , we do not change the sign. But , I want to know the reason behind it.

I mean if x<1 , x-1 is less than 1 , so we did change the sign. Like this I want to know the reason.

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we are not changing the sign bcz x-1 is less than 1. we are changing it bcoz say, x=0.9999, then 0.9999-1=-0.00001, which is a -ve value. after coming out of mod, it will be +ve.