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y= l x-1l+2
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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........
Posted by oldnick on May 30, 2013 at 1:54 AM (Answer #1)
I'm sorry I meant doesnt' esxists derivative:
Indeed: `D^+= lim_(h->0^+) (f(1+h)-f(1))/h= `
And `D^(-) = lim_(h->0^-) (f(1+h)-f(1))/h=lim_(h->0^-) (h+2-2)/h=1`
So `D^+ != D^-`
Posted by oldnick on May 30, 2013 at 2:10 AM (Answer #2)
We have given
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)+2` if `(x-1)<0`
Now we sketch graph (i) in red and (ii) in green.
Posted by pramodpandey on May 30, 2013 at 6:10 AM (Answer #3)
High School Teacher
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 ,
(ii) when x<1,
Posted by rakesh05 on May 30, 2013 at 6:44 AM (Answer #4)
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