Why cant I do this? (Regarding finding domain of function)

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The discussion focuses on determining the domain of a function defined by inequalities. It highlights the necessity for both inequalities, 9 - 4x ≥ 0 and 2x - 1 > 0, to be satisfied for the function to be defined. A key mistake identified is in the manipulation of the inequality, where subtracting 2 from both sides is incorrect; instead, dividing by 2 is necessary to isolate x properly. The importance of correctly handling inequality signs when dividing by negatives is also emphasized. Understanding these principles is crucial for accurately finding the domain of the function.
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Homework Statement



Problem:

http://archives.math.utk.edu/visual.calculus/0/domain.1/f6.gif

Homework Equations


The Attempt at a Solution



Look at the pic, and tell me again why I can't do this:

http://i47.tinypic.com/bhadsj.jpg
 
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-4x +4 is not x

Also be careful with the direction your inequality sign is supposed to point if you start dividing by negatives.
 
In addition to what aftershock said, you are missing something else. For f(x) to be defined, it must be true that 9 - 4x ≥ 0 AND 2x - 1 > 0. The domain for this function will be all of the numbers that satisfy both inequalities.

Also, you have a mistake in this work, similar to the one that aftershock pointed out.
2x - 1 > 0
2x > 1 (OK, so far)
Then you subtract 2 from both sides. What's wrong with that is that 2x - 2 ≠ x. To isolate x, divide both sides by 2.
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
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