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y= l x-1l+2

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y= l x-1l+2

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oldnick's profile pic

Posted (Answer #1)

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`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........

oldnick's profile pic

Posted (Answer #2)

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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^-`

pramodpandey's profile pic

Posted (Answer #3)

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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.

 

rakesh05's profile pic

Posted (Answer #4)

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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.

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