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mathmath333
 one year ago
State if \(f\) is a function
mathmath333
 one year ago
State if \(f\) is a function

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mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0A relation \(f\) is defined by \(\large \color{black}{\begin{align} f(x) = \begin{cases} x^2, & 0\leq x\leq 3 \\ 3x, & 3\leq x\leq 10 \end{cases}\hspace{.33em}\\~\\ \end{align}}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0How would you begin?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0(What are the conditions for f being a function? )

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Actually, it's a rule that for f being a function, there could only be 1 Yvalue per xvalue. So in every x, the function could be unexisting or it could have 1 yvalue.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So here you should look if there isn't a xvalue with two different yvalues

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0i didnt understand

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You should look, if for x=3, if there are 2 different yvalues. If there aren't two different values, it is a function

uri
 one year ago
Best ResponseYou've already chosen the best response.0@mathmath333 why are you faking as tania sachdev?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0In this case, you should only look at the case for x=3 because the two parts are both functions, the only problem could be, with the chosen intervals that for x=3 (in this case) the yvalue of the first function is different from the second

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0so it is same for x=3

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0If they're the same, you have just one Yvalue for the xvalue, s it's a function
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