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Dr.Peterson
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You want to solve the inequality [MATH]1\le h(x,y)\le 3[/MATH], that is, [MATH]1\le 3y-3\le 3[/MATH]. Can you do that?
It does get my final answer
Steven G
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I do not think so. Can you please post your answer?
The answer is : \[y\in [\frac{4}{3},2]\]
But i'm wondering why do we solve the inequality \[1\leq h(x,y) \leq 3\]
instead of \[1\leq h^{-1}(x,y) \leq 3\]
Dr.Peterson
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Well, what does h-1([1,3]) mean? It's the set of all points (x, y) in the domain such that f(x, y) is in [1, 3]. All I did was to express that symbolically.
In general, an inverse function amounts to the same thing as solving an equation.
But there's a very specific reason you can't use h-1(x,y): that doesn't even exist! Since the codomain of h is the real numbers, the domain of h-1 is the real numbers, not a set of ordered pairs. You'd have to talk about h-1(z), and for what values? I'm not sure you're really thinking.
I'm not positive what Jomo is saying, but you will certainly have to express your answer in terms of a set, not just a bare inequality, right?
In any case, it's not hard to solve, so when you have a solution, all you have to do is show us your answer to the problem and we'll tell you whether it's lacking anything.