How do I solve piecewise functions? I cannot tell you how many times I have stared at my notes and done the practice. https://gyazo.com/8740807019fb8b146a7ec27a3565554e

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How do I solve piecewise functions? I cannot tell you how many times I have stared at my notes and done the practice. https://gyazo.com/8740807019fb8b146a7ec27a3565554e

Mathematics
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f(-2) can be evaluated by the piece that includes x=-2
is -2 in the set of numbers that is described by x>=0? by x<-3? or by -3<=x<0?
Yes? the last one?

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yes the last one -2 is between -3 and 0
so you use f(x)=2 since x is between -3 and 0
which means f(anything in that interval)=2
f(-1)=2 f(-1/2)=2 f(-2.534)=2
so that means my answer is 2? So okay, the way I see is if the f(-2) or whatever number it is, happens to be in the set of functions, then the answer to f(x) is the opposite?
what?
I used f(x)=2 since x is between -3 and 0
if it said evaluate f(-5) I would have used f(x)=|x+3| since x is less than -3
f(-5)=|-5+3|=|-2|=2
so it has to fit in the thing? i'm confused now.
remember the first thing we looked at is what set of x values our x was included in then we used the corresponding piece
another example if we wanted to evaluated f(-10) -10 is less than -3 and our function says use f(x)=|x+3| if x is less than -3 so f(-10)=|-10+3|=-7|=7
another example if we wanted to evaluate f(20) well 20 is greater than 0 and our function says to use f(x)=x if x greater than or equal to 0 so f(20)=20
You lost me.
oh which part?
f(-10) or f(20) or what?
which means f(anything in that interval)=2
Like I said, I really don't get these piecewise functions..
before I said that I said this "so you use f(x)=2 since x is between -3 and 0" the interval being [-3,0) actually since it says to include -3 f(anything in that interval)=2 means any x value in [-3,0) will give us the output 2 when put into our function examples f(-3)=2 f(-2.9999)=2 f(-2.5)=2 f(-2)=2 f(-1.99)=2 f(-1.5)=2 f(-1/2)=2 f(-1/5)=2
but how did you get 2 from -2?
Your function says to use f(x)=2 if -3<=x<0
and -2 certainly falls in that interval
f(-2)=2 since -3<=-2<0
Okay, so that little comma thing is like "if this answer is right, use "2" for x "
no x is -2 in your problem f(x) or y is 2 for your problem
if it did say evaluate f(2) then you would look at the inequalities (or intervals whatever you want to call them) and see which one it satisfies 2>=0 so we use f(x)=x since x>=0 for this example so f(2)=2
does this still make no sense?
your function says: if x>=0 then use f(x)=x if x<-3 then use f(x)=|x+3| if -3<=x<0 then use f(x)=2
find which if part is true for your x then use function that corresponds to that inequality
to find the output for the function
x = -2 and y = 2
then what do I do with all these commas?
You don't need to do anything with them
if you want you can replace them with the word if if you prefer that
but you already found f(-2)
okay, then what do I do with it?
what does it refer to?
what third bit?
I'm absolutely lost on what we are talking about now
is there another question or something?
The only question I see is find f(-2) and we done that I don't see a third or even a second question on the picture you posted
but what did we find? I'm confused on what we actually found.
we found f(-2)
and what is f(-2)? 2?
yes f(-2)=2
remember if we have -3<=x<0 then we use f(x)=2 and we had x=-2 was included in -3<=x<0 so we used f(x)=2 to find f(-2)
insert any number from the inequality -3<=x<0 into f(x) and you get the output 2 insert any number from the inequality x>=0 into f(x) you get the output x insert any number from the inequality x<-3 into f(x) you get the output |x+3|
so my answer to this problem is 2, which I think is C?
So I guess you are asking me this question for the 4th time because you still don't understand why f(-2)=2?
pretty much.
Ok do you understand the below? \[f(x)=x \text{ if } x \ge 0 \\ f(x)=|x+3| \text{ if } x<-3 \\ f(x)=2 \text{ if } -3 \le x <0 \]
we plugged in -2 we are looking for f(-2) we replaced x with -2 replace all the x's with -2 \[f(-2)=-2 \text{ if } -2 \ge 0 \\ f(-2)=|-2+3| \text{ if } -2<-3 \\ f(-2)=2 \text{ if } -3 \le x <0\] the only line that is true here is the last line do you see why?
*\[f(-2)=-2 \text{ if } -2 \ge 0 \\ f(-2)=|-2+3| \text{ if } -2<-3 \\ f(-2)=2 \text{ if } -3 \le -2 <0\]*
the only line that is true here is the last line do you see why?
the last line is the only line that is true since: \[f(-2)=-2 \text{ if } -2 \cancel{\ge} 0 \\ f(-2)=|-2+3| \text{ if } -2\cancel{<}-3 \\ f(-2)=2 \text{ if } -3 \le -2 <0\]
yes, i get why we have 2. kinda. but i'm not sure what to do with it. Is it the answer to the problem?
the last line says f(-2)=2 ...
which makes 2 our answer. so the answer to the original problem is 2.
did you want to try more examples? if so see if you can find f(-100)

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