Thread: limit and graph

1>
Let f(x)=|x-1|+x-1
(a) sketch the graph of f(x)
(b) find lim x-> -2f(x)
(c) find lim x->2 f(x)

2> Find a cutoff distance δ such that if the distance from x to 1 is less than δ and x>1, then f(x) =3x-1. approximates 2 with an error of at most 1/100

Originally Posted by bobby77

1>
Let f(x)=|x-1|+x-1
(a) sketch the graph of f(x)
(b) find lim x-> -2f(x)
(c) find lim x->2 f(x)

The curve is below.
Note if your graphing software does not have the absolute value function you can graph y=sqrt(x^2) that is a nice trick to replace it.

To find the limits is very easy.
Note that,
f1=|x| is continous everywhere,
f2=x-1 is continous everywhere,
Thus,
f1 o f2=|x-1| is continous everywhere, (by composition rule).
Thus,
(f1 o f2)+f2=|x-1|+x-1 is continus everywhere, (by addition rule of continous functions).
Thus,
to find the limits of f=|x-1|+x-1
All we need to is evaluate the function at those points.

Originally Posted by bobby77

2> Find a cutoff distance δ such that if the distance from x to 1 is less than δ and x>1, then f(x) =3x-1. approximates 2 with an error of at most 1/100

Put x=1+x. x>0.

Then f(x)=2+3u, and if the error here is <=1/100 ,we have

f(x)-2<=1/100,

or:

3u<=1/100,

u<=1/300

So if 0<(x-1)<δ,and 0<=δ<=1/300 then f(x) =3x-1. approximates 2 with an error
of at most 1/100.

RonL