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11 Aug 2012, 19:39
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What is the best way to tackle a question of the following type?
Find the range of values of X if X^3 X^(1/3). I am trying to solve this using algebra.
Find the range of values of X if X^3 X^(1/3). I am trying to solve this using algebra.
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12 Aug 2012, 00:05
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sayak636
I prefer to get rid of the fractional exponents, therefore I use substitutions.
For the first inequality:
Denote \(x^{\frac{1}{3}}=y,\) so \(x=y^3.\) Then we have to find the values of \(y\) for which \(y^9<y\) or \(y(y^8-1)<0.\)
You have to check for values of \(y\) such that \(y<-1,\) \(\, -1<y<0,\) \(\, 0<y<1\) and \(y>1.\)
It follows that \(y<-1\) or \(0<y<1\). Translating back to \(x,\) we deduce that \(x<-1\) or \(0<x<1.\)
For the second inequality:
Denote \(x^{\frac{1}{3}}=y\), again \(x=y^3\). Then we have to find the values of \(y\) for which \(y^6>y\) or \(y(y^5-1)>0.\)
Now you have to test for values of \(y\) such that \(y<0,\) \(\, 0<y<1\) and \(y>1.\)
You find that \(y<0\) or \(y>1\). In terms of \(x\), this means \(x<0\) or \(x>1.\)
Look for most posts about inequalities on the site. Here is just one useful one:
solving-inequalities-134671.html
12 Aug 2012, 06:35
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Thank you. That makes good sense. I guess there is no direct way to solve equations such as (y^5 -1) or (y^8 -1).
