Inverse Function

[TrueTears]

Let

Find stating the rule, domain and range.

Basically i got the answer, but do we have to state the rule as a modulus function? Or can we just state the rule for the positive part of the modulus, since the negative is not applicable...?

how would you state the answer as a modulus anyway? i get a hybrid function like this



how do you turn that hybrid function into a modulus and restrict it for only the positive part?

thanks       

[Mao]

firstly, a few latex stuff,
use "\left[ \frac{1}{2},\infty \right)" to make the bracket the entire length of the fraction [looks funny the way you have it],
"\to" for the --> arrow,
"\mathbb{R}" for the set of real,
and you also need an "\end{cases}" at the end of your hybrid [otherwise typesetting cracks it and gives you wierd spacing]


You pretty much got everything right, though the convention isn't to write the answer as a hybrid. The answer is not a piecewise function, it rather has two branches. It cannot be expressed with a constraint in terms of x [because x can correspond to up to two y values]. In this case, you have to restrain the domain of the original function so it is one-to-one.

Also, you cannot graph this as a modulus. sketch the graph and compare to the graph of a modulus, they are not the same. =]

[TrueTears]

oh yeah thanks mao, so what would i put as the answer to this Q...?

[Mao]

try doing this: draw the first function, note the domain stated in the question, that may shed some light on which part you should include in your answer.

[TrueTears]

thanks mao i got it now hehe