# y= l x-1l+2

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given function is y=mod(x-1)+2.

We know that mod(x)=x if x>0

and =-x if x<0.

So here mod(x-1)= x-1 if x>1

=1-x if x<1.

So (i) When x>1 ,

y=mod(x-1)+2

=(x-1)+2

=x+1.

(ii) when x<1,

y=mod(x-1)+2

=1-x+2

=3-x.

We have given

`y=|x-1|+2`

Though graph is simple ,except problem is at x >1 and x<1.

So let define

`y=x-1+2 ` if `(x-1)>=0`

```y=x+1` (i)

and

`y=-(x-1)+2` if `(x-1)<0`

`y=-x+3` (ii)

Now we sketch graph (i) in red and (ii) in green.

I'm sorry I meant doesnt' esxists derivative:

Indeed: `D^+= lim_(h->0^+) (f(1+h)-f(1))/h= `

`=lim_(h->0^+) (2h-2)/h=-1`

And `D^(-) = lim_(h->0^-) (f(1+h)-f(1))/h=lim_(h->0^-) (h+2-2)/h=1`

So `D^+ != D^-`

`y=|x-1|+2`

This function is always positive,definte `AA x in RR` and so is the range.

`|x-1|=` `1-x` if `x<1` and `x-1` if `x>1`

so `|x-1| +2=` `3-x` if `x<1` and `x+1 ` if `x>1`

This funtion is continue even tought in the point `x=1` doesn't exists limit.

TO BE CONTINUE........