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#Post#: 93--------------------------------------------------
Re: Unit 3 forum
DIR By: dluvbell
Date: November 4, 2014, 6:57 pm
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--- Quote from: idk link ---
>
> Determine the intervals where 2
> x+1 > x
>
--- End Quote ---
I assume the question is (x+1) > (2/x).
Remember....Always move everything to one side. So it becomes
x+1 - (2/x) > 0 if you move (2/x) to the left side.
Then you combine x+1 and 2/x by finding common denominator like
you were computing addition/subtraction of fractions.
So it becomes x(x+1)/x - 2/x >0.
So then [x(x+1)-2]/2 >0.
From this, you know what to do ;)
#Post#: 94--------------------------------------------------
Re: Unit 3 forum
DIR By: Thomath
Date: November 4, 2014, 8:05 pm
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why h(x)=(2x-1)/(x+2) has a branch in second quadrant? dose it
has anything to do with the sign of the whole function？
but there is no negative sign~~~
#Post#: 95--------------------------------------------------
Re: Unit 3 forum
DIR By: dadougiedude
Date: November 4, 2014, 8:15 pm
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it doesn't matter if there isn't negative sign, for the interval
-2<x<0.5 it's under x-axis, which means f(x)<0
in short, you must do the number line table to find the end
behaviour for rational functions. (so you can graph the function
fully, not just the knowing the end behaviour)
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