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# Suppose the domain of f is (-1,1). Define the function l by l(x)= f(x+1/x-1). What is the domain of l?

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Suppose the domain of f is (-1,1). Define the function l by l(x)= f(x+1/x-1). What is the domain of l?

waffles  Oct 25, 2017
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#1
+5254
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The domain of  f(x)  is  -1 < x < 1

So the domain of  f( $$\frac{x+1}{x-1}$$ )  is  -1 < $$\frac{x+1}{x-1}$$ < 1

*edit* (I was unhappy with my previous answer.)

The easiest way to find the  x  values that make this inequality true is to look at a graph.

So the  x  values that make  -1 < $$\frac{x+1}{x-1}$$ < 1  true are  x < 0 .

The domain of  f( $$\frac{x+1}{x-1}$$ )  is  (-∞ , 0)

The domain of  l(x)  is  (-∞ , 0) .

hectictar  Oct 25, 2017
edited by hectictar  Oct 28, 2017
#2
+78744
+1

One question, hectictar...

You write that

" So the  x  values that make  -1 <  (x +1) / (x - 1)  < 1  true will either be  x < 0  or  x > 0 ."

How did you determine this  ???

CPhill  Oct 25, 2017
#3
+5254
+1

x + 5  <  8

If for some reason we cant solve this inequality the normal way, then we can do it like this

x + 5  =  8

x  =  3

And then we know that

the values that make the original inequality,  x + 5 < 8 , true will be either  x < 3  or  x > 3 .

I thought about it because when you are trying to find the values that make a quadratic function greater or less than zero, we have to do it like this and test points in an interval.

....Is that a good explanation??

hectictar  Oct 25, 2017
edited by hectictar  Oct 25, 2017
edited by hectictar  Oct 25, 2017
edited by hectictar  Oct 28, 2017
#4
+78744
+1

Got it.....thanks.....!!!!!

CPhill  Oct 25, 2017

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