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- dancing_stars23

Find f(-2) for the following piecewise function.... medal

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- dancing_stars23

Find f(-2) for the following piecewise function.... medal

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- dancing_stars23

|dw:1444177967113:dw|

- candyme

First you need to figure out which equation to plug the number in.
If the number is greater than or equal to 0, use the first one
If the number is less than -3, use the second one
If the number is greater (or equal to) -3 but still less than 0, use the last one.
So where does -2 fit?

- dancing_stars23

the third one then
@candyme

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- dancing_stars23

right?i am confused....

- candyme

Yep. Since there isn't a place to plug in the variable in for the third equation, whatever is there is your answer. So your answer is f(-2)=2

- candyme

Piecewise functions can be a little tricky at first, but you got it! Let me know if I can help with any more :)

- dancing_stars23

so every time i have a problem i will use the section where it says find f({any number}). What does the number(s) before each comma in the equation mean

- dancing_stars23

it means the answer I am guessing?

- dancing_stars23

I really don't know. I just learned about piecewise but about graphing not equations like these

- candyme

The parts before the comma are the equation you put the variable into. So if you were looking for f(-4) for example, you would use the equation |x+3| and plug in -4 for x
so you would then get |-4+3|=|-1|=1 for your answer.

- candyme

If there is an x in the part before the comma, then you plug in the number you are trying to find
If there is no x in the part before the comma, then that is your answer

- dancing_stars23

okay, that makes perfect sense now! thanks so much! I am doing online schooling and it is so much harder to learn math on there then it is in a classroom setting!!!

- candyme

I totally get that. I actually did algebra 1 and 2 online and it was much more difficult. OpenStudy was a lifesaver because you can get explanations. I have used it ever since then. You are definitely in the right place!

- dancing_stars23

@candyme thanks so much for your help! :)

- candyme

No problem! Good luck with your class! I will see you around the site :)

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