<p>Each of the following inequalities is true for some values of x EXCEPT</p>
<p>A) x < x^2 < x^3</p>
<p>B) x < x^3 < x^2</p>
<p>C) x^2 < x^3 < x</p>
<p>D) x^3 < x < x^2</p>
<p>E) x^3 < x^2 < x</p>
<p>Okay, so I found certain values for x and found that A, D, and E would be impossible. </p>
<p>But, I'm stuck between B and C. </p>
<p>Which would be an example that would make B or C impossible? </p>
<p>Thanks in advance!</p>
<p>If you aren’t familiar with exponents and their rules, then you’ll have a bit of trouble with this question. Since I don’t know them myself, I’ll tell you how I would have approached the question.</p>
<p>I would raise the following numbers: -2, -0.5, 0.5, 2 to the second, and third powers, and arrange the three numbers in ascending and descending order, then replace the number with “X”. Obviously, each number would result in a different inequality, then I’d pick off the odd one out.</p>
<p>-2^2 = 4, -2^3 = -8. Therefore, x^3 < x < x^2. Choice D checks out.
And so on with the other three numbers.</p>
<p>B can be true:
if x = -1/2
x = -1/2
x^3 = -1/8
x^2 = 1/4</p>
<p>Therefore, that leaves C.</p>
<p>You should also know that x^3 is only greater than x^2 when x is a negative number. However, because x is greater than x^2, we know x is not a negative number (if it was x^2 would be positive).</p>
<p>Why C ( x^2 < x^3 < x ) is impossible:</p>
<p>Suppose x is positive. Divide the left inequality by x to get x < x^2. But, if you ignore the middle part, C tells us that x^2 < x. So that can’t be right.</p>
<p>Suppose x is negative. Then it’s impossible since x^2 is positive and x is negative.</p>
<p>Thank you all so much! </p>
<p>More questions to come soon, hehe!</p>
<p>@SirWanksalot</p>
<p>-2^2 = 4 is FALSE</p>
<p>Exponentiation is always done before multiplication:</p>
<p>-2^2=(-1)(2^2)=(-1)(4)=-4</p>
<p>What you meant to say was: </p>
<p>(-2)^2=4</p>
<p>Same goes for -2^3.</p>
<p>Yeah, you’re right. I meant -2^2 as in (-2)^2.</p>