Calculate the following:

Question

a) Find g(f(1))
b) Find lim x-> 1+ g(f(x))
c) Find f(g(0))
d) Find lim x->0- f(g(x))

f(x)= 1-x, x<1
1, x=1
x-1, x>1

g(x)= -x, x<0
2, x=0
x+2, x>0

This is where it gets hard. I have a graph for f(x) and one for g(x). I will explain them. The graph for f(x)has an open circle at 1 on the x-axis and a line extending straight out from it going to the right. Another line extends out from it as well in the left direction and passes through (0,1). A closed circle is located at (1,1).

The graph for g(x) has an open circle at the origin with a line extending from it going left. A closed circle is located at (0,2) with a line extending from it going to the right.

I have been struggling with these and I am now also confused as to why he included the info with the x is <,=,> part, as I have never seen this before. Could someone please help. I would really appreciate it.

Answer 1

f(x) and g(x) are discontinuous functions defined each by two lines of different slopes and a point at the discontinuity, exactly as you described it.

Before finding f(g(x)), you will need to find out the different intervals on which both f(x) and g(x) are continuous, for example,
1. x<0
2. x=0
3. 0<x<1
4. x=1
5. x>1

You can then define f(g(x)) on each interval, similar to the way the original functions are defined.

For example, for interval 1, (x<0),
f(x)=1-x, g(x)=-x
so f(g(x))=f(-x)=1-(-x)=1+x
for interval 2, (x=0)
f(x)=1, g(x)=2
so f(g(x))=f(2)=1
and so on.

For the limits, you will have to consider the values of f(x) or g(x) approaching from either side of the limit, for example, x=1+ or x=1-.

Answer 2

I'm confused. Why do you use f(x)= 1-x, g(x)=-x... I don't see that interval, but I see that interval 2 is the 2nd line for g(x).

???

Answer 3

I'm just as confused!

Answer 4

For x<0, f(x) is defined as 1-x, and g(x) is defined as -x.
The interval (-∞,0) is created by g(x), because g(x) has a continuity at x=0.

Post if you have other questions.