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I don't understand this question at all. For (a) and (b), the two equations are on separate lines in a curly bracket - I wasn't sure how to format this so I just separated them using a semi-colon instead.

I know what all of these look like graphed, I am just not sure what defines a function (other than if the $x$ values have multiple $y$ values.

Adam Rainey
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  • I think you got the definition of function right (although you're formulating it the other way around): One x-value may only lead to one y-value. That's the only thing necessary to solve the questions. Hint: Watch out for the inequality signs! – Lisa Oct 03 '13 at 22:13
  • To see how to format a function defined by cases, right click and Show Math As -> TeX Commands. The double backslash gives a line break, the & gives the spacing between columns – Ross Millikan Oct 03 '13 at 22:14

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(b) is not a valid function since $f\left(0\right)=0$ by the first specification and also $f\left(0\right)=1$ by the second specification (contradiction, for each $x$, $f\left(x\right)$ should be unique)

you could argue that (c) is not a valid function if you are assuming $f$ is a function from the real numbers (e.g. $3$, $1/2$, $\pi$, etc.) to the real numbers as $1/\left(1-1^2\right)$ is undefined in that case (division by zero). I think from the context, this is what the author of the question wants you to assume.

parsiad
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