Since, we were all having so much fun with inequalities.. I

Question

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.

banerjeea_98 · Answer

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

MA · Answer

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?

MA · Answer

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.

cloaked_vessel · Answer

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?

thearch · Answer

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?

banerjeea_98 · Answer

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

thearch · Answer

yes, I get it now. Thanks

Since, we were all having so much fun with inequalities.. I

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