Question

Need help with this problem...
I believe that there are discontinuities at x=2, x=1, and x=0

Answer 1

What do they want me to write for the properties?

Answer 2

This is my last question and I really want to finish.

Answer 3

Okay, I'll give my best to help you.
Did they give you a function or just the graph?

Answer 4

Just the graph

Answer 5

Okay, let's first begin by studying the domain:
Observing the graph, we can see that the domain are all the values excepting the 1 value.
you can write it like this: D(f)=R-(1)

Answer 6

We have a lot of jumps here, so that means that the function has some values that jump.
, so let's analyze the limits of the given points:
\[\lim_{x \rightarrow 0^{-}}f(x)=0\]
same way:
\[\lim_{x \rightarrow 0^{+}}f(x)=0\]
When we talk about "0" in limits, we say that we take values, extremely small, so with that I mean that I never touch zero, so the limits will approach zero but the image of zero is 1.

Answer 7

Try using that conception on the value "2"

Answer 8

Limit as x approaches 2 from the negative side of f(x)= -1
Limit as x approaches 2 from the positive side of f(x)= -1

Answer 9

good! Now try it for the value of 1.

Answer 10

The limit as x approaches 1 from the positive side f(x)= Positive infinity
The limit as x approaches 1 from the negative side f(x)=Negative infinity.

Answer 11

And with that we finished the study of limits, but pluging in the actual values on the function, taking for example the value 1.
We can see that f(1) = 0 and do that on all the pints we analyzed.

Answer 12

So instead of saying as x approaches 2 figure out what the number is at that point

Answer 13

I am not sure what f(0) is.

Answer 14

is it 1?

Answer 15

f(2) =1
f(1) = 0
f(0)= 1 or 0

Answer 16

Now, instead of actually approaching the numbers, we are analyzing what happen in the actual numbers,.
I believe it's a error of the person who made the problem, but no worries, teachers are also human hehe.
And with that we have analyzed the function.

Answer 17

Do you know what f(0) is though?

Answer 18

or a guess to what it is?

Answer 19

Thank you for helping me. I am just going to say f(0)=0

Answer 20

I'm not sure, but yeah.
No problem :)