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Manager
Joined: 14 Dec 2004
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Since, we were all having so much fun with inequalities.. I [#permalink]
 14 Apr 2005, 16:48
Since, we were all having so much fun with inequalities.. I looked into my old high school algebra book and pulled one out...
Solve:
x^(-3) - x^(-2) <= 0
Also, can you state how long it took to arrive at the solution. I have to work on my speed for inequality computations.
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SVP
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1/x3 <= 1/x2
multiply by x^4
==> x <= x2
x^2 - x >= 0
x(x-1) >= 0
solution is X >=1 or X < 0
10 secs
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SVP
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x^(-3) - x^(-2) <= 0
1/(x^3) - 1/(x^2) <= 0
1/(x^3)<= 1/(x^2)
x^2 <= x^3
x^3 - x^2 >= 0
x^2(x-1)>= 0
x >= 0, or x>=1
arround 20 seconds.
But, let me ask you gus, why x is not >=0? i know it does not fit to the inequality, but how is it?
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MA wrote: But, let me ask you gus, why x is not >=0? i know it does not fit to the inequality, but how is it?
ok, i got it. x^2(x-1)>= 0
x^2 >= 0, x >=0 , x<=0,
x cannot be >=o, x can only be <=0.
therefore, x>=1, x<=0.
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Manager
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I don't think its legal to multiply by x^3 in an inequality.
I think MA you have done that in arriving at your solution.
Because you don't know if the sign will change or not. I am concluding this from the thread on number properties:
"x>1/x
Again you CAN'T multiply both sides by x because you don't know if x is positive or negative. What you have to do is to move the right side to the left:
x-1/x>0
"
I think multiplying by x^2 is ok.
Is this notion correct?
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Director
Joined: 19 Feb 2005
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I'd add another question: how could you multiply the 2 members of the inequality by x^4 if x could be = 0?
I solved this way:
1/x^3-1/x^2 <= 0
1/x^2(1/x-1)<=0
since 1/x^2 is always positive, the sign is determined by 1/x-1
so 1/x<=1 => x>=1 (I lost the "x<0" solution)
Where is the mistake?
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thearch wrote: I'd add another question: how could you multiply the 2 members of the inequality by x^4 if x could be = 0?
I solved this way:
1/x^3-1/x^2 <= 0
1/x^2(1/x-1)<=0
since 1/x^2 is always positive, the sign is determined by 1/x-1
so 1/x<=1 => x>=1 (I lost the "x<0" solution)
Where is the mistake?
x can't be = 0....as we have terms in 1/x, which will make 1/x = infinity...in ur soln, everything is correct except when u reverse
1/x <= 1....u can't just say x > = 1....consider when x < 0, that also satisfies 1/x <= 1
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Director
Joined: 19 Feb 2005
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yes, I get it now. Thanks
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close
